Demo 8

Trigonometry 101 . 02. Trigonometry (I) Problems based on Trigonometric Ratios 1. If tan3θ.tan7θ = 1, where 7θ θ θ θ is an acute angle, then find the value of cot15θ. θ. θ. θ. (a) 3 − (b) 1 (c) –1 (d) 3 SSC MTS 03/05/2023 (Shift II nd ) Ans. (c) : tan3θ.tan7θ = 1 (where7θ is an acute angle) tan7θ = cot3θ tan7θ = tan(90° – 3θ) 7θ = 90° – 3θ then (3θ + 7θ) = 90° 10θ = 90° θ = 9° then cot 15θ cot (15×9)° = cot 135° = cot (90×1+45°) = – tan 45° = –1 2. If sinθ = θ = θ = θ = 8 17 then find the value of tanθ. (a) 17 15 (b) 8 15 (c) 15 17 (d) 15 8 SSC CGL 05/12/2022 (Shift-I) Ans. (b) : Perpendicular sin Hypotenuse θ= sinθ = 8 17 AB = 2 2 17 8 − = 289 64 − = 225 = 15 P tan B θ= Hence tanθ = BC 8 AB 15 = 3. If cosA = 9 41 , then find the value of cotA. (a) 41 40 (b) 9 40 (c) 40 9 (d) 9 41 SSC CGL 01/12/2022 (Shift-IV) Ans. (b) : Given that 9 cos A 41 = Base cos A Hypotenuse   =     Q 2 2 BC 41 9 = − 1681 81 40 = − = Base cot A Perpendicular = 9 40 = 4. If cotA = 15 8 then get the value of tan2A will be? (a) 240/173 (b) 240/161 (c) 220/171 (d) 200/161 SSC CGL 06/12/2022 (Shift-II) Ans. (b) : According to question, 15 cot A 8 = then 8 tan A 15 = 1 cot A tan A = 2 2 tan A tan 2A 1 tan A = − 8 2 15 64 1 225 × = − 16 225 240 15 161 161 × = = × 5. If 8cotθ = 6 then find the value of sinθ + cosθ sinθ - cosθ . (a) 7 (b) 2 (c) 5 (d) 12 SSC MTS 01/09/2023 (Shift I st )

Trigonometry 102 . Ans. (a) : 8 cotθ = 6 B cot P θ= 6 8 = AC 2 = 6 2 + 8 2 AC 100 = AC = 10 Hence, sin cos sin cos θ+ θ θ− θ 8 6 10 10 8 6 10 10 + = − 14 10 10 2 = × = 7 6. ∆ABC is a Right angle triangle. If ∠B = 90° and tanA = 1 3 then find the value of (sinA cosC + cosA.sinC). (a) 1 3 (b) 2 3 (c) 2 (d) 1 SSC CGL (Tier-I) 26/07/2023 (Shift-II) Ans. (d) : tanA = 1 3 tanA = tan30º A = 30º ∠B = 90º then ∠C = 180º – (∠A + ∠B) = 180º – (90º + 30º) = 60º sinA.cosC+cosA.sinC = sin30º.cos60º+cos30º.sin60º 1 1 3 3 1 3 2 2 2 2 4 4 = × + × = + 1 3 1 4 + = = 7. If tanA = 5 12 then the value of cosA is : (a) 5 13 (b) 13 5 (c) 12 13 (d) 12 5 SSC CGL 05/12/2022 (Shift-IV) Ans. (c) : 5 tan A 12 = (given that) AC 2 = AB 2 + BC 2 = (12) 2 + (5) 2 144 25 = + 169 = AC = 13 then B cos A H = 12 13 = 8. If secθ – 2cosθ = 7 2 , where θ is positive acute angle, then the value of secθ is. (a) 5 (b) 6 (c) 4 (d) 8 SSC CGL 06/12/2022 (Shift-IV) Ans. (c) : 7 sec 2cos 2 θ− θ= ⇒ 7 sec 2 cos 2 θ− θ= 1 sec cos   θ=   θ   Q ⇒ 1 7 2 cos cos 2 − θ= θ ⇒ 2 1 2 cos 7 cos 2 − θ = θ ⇒ 2 2 4 cos 7 cos − θ= θ ⇒ 2 4 cos 7 cos 2 0 θ+ θ− = ⇒ 4cos 2 θ + 8cosθ – cosθ – 2 = 0 ⇒ ( ) ( ) 4 cos cos 2 1 cos 2 0 θ θ+ − θ+ = ⇒ ( )( ) cos 2 4 cos 1 0 θ+ θ− = 4 cos 1 0 θ− = 1 cos 4 θ= ∴ secθ = 4 9. Find the accurate value of cos 120º (a) 1 (b) –0.5 (c) 0 (d) 0.5 SSC CGL (Mains) 06/03/2023 Ans. (b) : According to question, cos 120º = cos (180º – 60º) = – cos 60º = 1 2 −       = – 0.5 10. If sinθ = 9 41 , and 0° <θ<90°, then the value of cotθ is : (a) 39/9 (b) 35/8 (c) 40/9 (d) 47/8 SSC CGL (Tier-II) 08/08/2022 (Shift-I)

Trigonometry 103 . Ans. (c) : Given that 0 0 9 sin ,0 90 41 θ= <θ< 2 cos 1 sin θ= − θ 2 9 1 41   = −     1681 81 40 1681 41 − = = ∴ 40 cos 41 cot 9 sin 41 θ θ= = θ 40 9 = 11. In right-angled triangle PQR, PQ = 5cm, QR = 13 cm and ∠P = 90º then find the value of (tanQ – tan R). (a) 60 119 (b) 119 60 (c) 14 5 (d) 5 14 SSC CHSL (Tier-I) 17/08/2023 (Shift-II) Ans. (b) : According to the question In right-angle triangle QPR, By Pythagoras Theorem, 2 2 PR 13 5 = − = 12 cm tanQ – tanR 12 5 5 12 = − 144 25 60 − = 119 60 = 12. Suppose ∆ABC is a right-angled triangle where ∠A = 90° and ∠C = 45º, then find the value of (secC + sinC.secC). (a) 1 2 + (b) 1 (c) 2–1 (d) 1– 2 SSC CHSL (Tier-I) 11/08/2023 (Shift-I) Ans. (a) : sec C + sin C sec C = sec45º + sin45ºsec45º = 1 2 2 2 + × 2 1 = + Ùee 1 2 + 13. If 3tanθ = 3sinθ , then the value of (sin 2 θ – cos 2 θ) is : (a) 1/2 (b) 1/3 (c) 1/5 (d) 1/4 SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (b) : Given that, 3 tan 3sin θ= θ sin 3 3sin cos θ = θ θ sin 3 sin cos θ = θ θ 1 cos 3 θ= 2 sin 1 cos θ= − θ 1 1 3 = − 2 3 = ∴ 2 2 2 2 2 1 sin cos 3 3     θ− θ= −           2 1 3 3 = − 1 3 = 14. If secA = 17 8 while A < 90° then, find the value of following expression 34sinA + 15cotA 68cosA - 16tanA . (a) 23 (b) 19 (c) 30 (d) 38 SSC CGL 11/04/2022 (Shift-III) Ans. (b) : 0 17 sec A , A 90 8 = < 8 cos A 17 = 2 2 8 15 sin A 1 cos A 1 17 17   = − = − =     sin A 15/17 15 tan A cos A 8/17 8 = = = 8 cot A 15 =

Trigonometry 104 . 15 8 34 15 34sin A 15cot A 17 15 8 15 68cosA 16 tan A 68 16 17 8 × + × + = − × − × 30 8 38 19 32 30 2 + = = = − 15. If 0 0 3tanθ = 2 3sinθ, 0 < θ < 90 then the value of 2 2 2 2 cosec 2θ + cot 2θ sin θ + tan 2θ will be : (a) 20 27 (b) 7 13 (c) 20 39 (d) 4 3 SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (c) : 3 tan 2 3 sin θ= θ sin 3 2 3 sin cos θ = θ θ 0 3 cos cos 30 2 θ= = 0 30 θ= 2 2 2 2 cosec 2 + cot 2 sin + tan 2 θ θ θ θ 2 0 2 0 2 0 2 0 cos ec 60 cot 60 sin 30 tan 60 + = + ( ) 2 2 2 2 2 1 3 3 1 3 2     +         =   +     4 1 20 3 3 1 3 39 4 1 + = = + 16. If 5 sinA = 13 and 7cotB = 24 then the value of (secA.cosB)(cosecB.tanA) will be : (a) 13 14 (b) 65 42 (c) 13 7 (d) 15 13 SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (b) : 5 sin A 13 = and 7 cot B = 24 For Angle (A) – P = 5 H = 13 By pythagoras theorem B 2 = H 2 – P 2 = 13 2 – 5 2 = 169 – 25 B 144 = B = 12 cot B = 24 7 = B P By Pythagoras Theorem, H 2 = B 2 + P 2 = (24) 2 + (7) 2 = 576 + 49 H 625 = = 25 (secA. cosB) (cosecB tanA) 13 24 25 5 12 25 7 12   = × × ×     65 42 = 17. In a right angled triangle ABC, where ∠C = 90° and sinA = sinB then find the value of cosA. (a) 3 2 (b) 1 (c) 1 2 (d) 1 2 SSC MTS 08/09/2023 (Shift I st ) Ans. (c) : According to the question, ∠C = 90°, sin A = sin B Q A + B + C = 180° 2A + C = 180° 2A = 180°–90° 2A = 90° A = 45° then, sin A = sin45° = 1 2 ∴ cos A = 2 1 sin A − cos A = 2 1 1 2   −     cos A = 1 2 Dele: efJekeâuhe (c) mener Gòej nw~ 18. If Y = tan35° then find the value of (2tan 55° + cot55°). (a) 2 2 Y Y − (b) 2 2 Y Y + (c) 2 2 Y (d) 2 2 Y Y − SSC MTS 09/05/2023 (Shift I st ) Ans. (b) : 2 tan 55° + cot 55° = 2 tan (90 o –35º) + cot (90 o – 35°) = 2 cot 35° + tan 35° o o tan (90 ) cot cot (90 ) tan   −θ = θ     −θ = θ     Q = 2 tan 35 tan 35 + ° ° = 2 Y Y + = 2 2 Y Y +

Trigonometry 105 . 19. If cosθ = 2 2 P –1 P +1 , 0° < θ < 90° then find the value of cosecθ. (a) 2 1 P 2P + (b) 2 1 P 2P − (c) 2 2P 1 P − (d) 2 2P 1 P + SSC MTS 03/05/2023 (Shift II nd ) Ans. (a) : Given that cosθ = 2 2 P 1 P 1 − + 0°< θ < 90° cosec θ = ? By Pythagoras Theorem Perpendicular 2 + Base 2 = (Hypotenuse) 2 x 2 +(P 2 –1) 2 = (P 2 +1) 2 x 2 +P 4 +1–2P 2 = P 4 +1+2P 2 x 2 =P 4 +1+2P 2 –P 4 –1+2P 2 x 2 = 4P 2 x 2 = (2P) 2 x = 2P ∴ cosecθ = Hypotenuse Perpendicular = 2 P 1 2P + 20. In ∆ABC, B is right angled and cot A = 1 2 then find the value of sinA(cosC + cosA) cosC(sinC - sinA) . (a) 3 (b) –3 (c) 2 (d) –2 SSC CPO SI 11/11/2022 (Shift III rd ) Ans. (b) : cotA = 1 2 = Base Perpendicular AC 4 1 5 = + = 2 1 1 2 sin A ,cosA ,sinC ,cosC 5 5 5 5 = = = = 2 2 1 2 3 sin A(cosC cosA) 5 5 5 5 5 3 2 1 cosC(sinC sinA) 2 1 2 5 5 5 5 5   × + ×   +   = = =− − −   × −     21. If 5 sinθ – 4 cosθ = 0, 0° < θ < 90°, then the value of 5sinθ - cosθ 5sinθ + 3cosθ is (a) 2/7 (b) 3/7 (c) 6/7 (d) 4/7 SSC CGL (Tier-I) 13/04/2022 (Shift-II) Ans : (b) Given, 5 sinθ – 4cosθ = 0, 0º < θ < 90º 5sinθ = 4 cosθ According to the question, 5sin cos 5sin 3cos θ− θ θ+ θ = 4 cos cos 4 cos 3cos θ− θ θ+ θ 3cos 7 cos θ = θ 3 7 = 22. If cos B 5 = , 7 What is the value of cosec B+cot B? Given that 0 < B < π 2 (a) 5 6 (b) 6 12 (c) 7 6 (d) 6 SSC CGL (Tier-I) 19/04/2022 (Shift-II) Ans. (d) Given 5 CosB 7 = 2 SinB 1 cos B = − 2 5 1 7   = −     24 49 = 2 6 7 = ∴ 1 cosB CosecB + cotB = + sinB sinB 7 5/7 2 6 2 6 7 = + 7 5 2 6 2 6 = + 12 2 6 = 6 =

Trigonometry 106 . 23. If 5 sinθ – 4 cosθ= 0, 0°<θ<90°, then the value of 5sinθ + 2cosθ 5sinθ + 3cosθ is: (a) 4/7 (b) 6/7 (c) 2/7 (d) 3/7 SSC CGL (Tier-I) 13/04/2022 (Shift-I) Ans. (b) Given, 5sinθ = 4cosθ According to the question, 5sin 2 cos ? 5sin 3cos θ+ θ = θ+ θ 4cos 2cos 6 cos 4 cos 3cos 7 cos θ+ θ θ = = θ+ θ θ 6 7 = 24. If tan 2 A + 2tan A – 63 = 0 Given that 0 < A < 2 π what is the value of (2sinA + 5 cosA)? (a) 19 50 (b) 15 50 (c) 19 50 (d) 15 50 SSC CGL (Tier-I) 11/04/2022 (Shift-III) Ans. (c) tan 2 A + 2tan A – 63 = 0 tan 2 A + 9tan A – 7tan A – 63 = 0 tan A (tan A + 9) – 7 (tan A + 9) = 0 (tan A + 9) (tan A –7) = 0 tan A = 7 Value of AC = 2 2 7 1 50 + = 2 Sin A + 5 CosA = 7 1 2 5 50 50 × + × = 19 50 25. If ≠ cotθ + cosθ k+1 = ,k 1, cotθ - cosθ 1-k then k is equal to: (a) sin θ (b) cosec θ (c) cos θ (d) sec θ SSC CHSL 09/08/2021 (Shift-I) Ans. (a) : cotθ + cosθ k+1 = cotθ cosθ 1 k - - 1 sin 1 k 1 sin 1 k + θ + = − θ − 1–k + sin θ – k sin θ = 1 + k – sin θ – k sinθ 2k = 2 sin θ k = sin θ Trick:– cot cos k 1 cot cos 1 k θ+ θ + = θ− θ − By componendo and dividendo rule cos k cot 1 θ = θ sin θ = k 26. If cosθ = 2 3 , then 2sec 2 θ + 2tan 2 θ –6 equals: (a) 1 (b) 4 (c) 2 (d) 0 SSC CHSL 15/04/2021 (Shift-I) Ans. (a) : Base 2 Given,cos Hypotenuse 3 θ= = By Pythagoras theorem, (Hypotenuse) 2 = (Base) 2 + (Perpendicular) 2 ( ) () () 2 2 2 Perpendicular 3 2 = − Perpendicular 9 4 = − 5 = According to the question, 2 2 2sec 2 tan 6 θ+ θ− H P sec , tan B B   θ= θ=     Q 2 2 3 5 2 2 6 2 2     = + −           9 5 2 2 6 4 4 = × + × − 9 5 6 2 2 = + − = 7 – 6 = 1 27. In ∆ABC, right angled at B, if cot 1 A= 2 , then the value of ( ) ( ) sinA cosC + cosA cosC sinC - sinA is : (a) 3 (b) –3 (c) –2 (d) 2 SSC CGL–(Tier-I) 24/08/2021 (Shift I) Ans. (b) : 1 B cot A 2 P = = ------------[Given]

Trigonometry 107 . By Pythagoras theorem, 2 2 H 1 2 5 = + = ( ) ( ) 2 2 1 sin A cosC cos A 5 5 5 2 1 2 cosC sinC sin A 5 5 5   +   +   = −   −     3 5 3 1 5 = =− − 28. If 5 sinA = 13 and 7 cotB = 24, then the value of (secA cosB) (cosec B tan A) is: (a) 65 42 (b) 13 14 (c) 15 13 (d) 13 7 SSC CGL (Tier-II) 29/01/2022 Ans. (a) According to the question, 2 2 AB 13 5 = − 2 2 AB 24 7 = + AB 144 = AB 625 = AB 12 = AB 25 = (secA.cosB) (cosecB tanA) 13 24 25 5 . . 12 25 7 12    =       65 42 = 29. In ∆PQR, ∠Q = 90 0 . If tan 1 R= , 3 then what is the value of ( ) ( ) secP cosR + sinP cosecR sinR - cosecP ? (a) 18 7 (b) 18 7 − (c) 2 7 − (d) 2 7 SSC CGL–(Tier-I) 18/08/2021 (Shift II) Ans. (b) : Given, ∠Q = 90º ] 1 P tan R 3 B = = From Pythagoras therom, 2 2 H P B = + 2 2 H 1 3 = + H 10 = Then, ( ) ( ) secP cosR+sinP cosecR sinR-cosecP 10 3 3 1 10 10 10 1 10 1 3 10   +     =   −       6 10 10 3 10 10 3 10 × = −       6 7 3 − 18 7 =− 30. If 1 sinA = ,A 2 is an acute angle, then find the value of ( ) tanA - cotA 3 1 + cosec A : (a) 4 3 9 (b) 2 9 (c) 2 9 − (d) 4 3 9 − SSC CGL–(Tier-I) 17/08/2021 (Shift II) Ans. (c) : Given that, 1 sin A 2 = sin A sin 30º = A 30º = Q On putting the value of A, ( ) ( ) ( ) 1 3 tanA cotA tan30º cot30º 3 3 1 cosecA 3 1 cosec30º 312 − − − = = + + + 1 3 2 3 9 33 − − = = 31. If sin θ = 11 15 , then the value of (secθ–tanθ) is: (a) 2 26 13 (b) 26 13 (c) 4 26 (d) 1 26 SSC CHSL 19/04/2021 (Shift-I)

Trigonometry 108 . Ans. (b) : sin θ = 11 15 (Given) 2 cos 1 sin θ= − θ = 121 104 4 26 2 1 26 225 225 225 15 × − = = = × So, (secθ–tanθ) = ( ) 11 15 15 2 26 2 26 15 −         = 15 11 2 26 2 26 − = 4 2 26 = 2 26 = 26 13 32. If 4 tanθ = , 3 then the value of 9sinθ + 12cosθ 27cosθ - 20sinθ will be equal to: (a) 72 (b) 36 (c) 18 (d) 100 SSC CHSL 10/08/2021 (Shift-I) Ans. (a) : tan θ = 4 3 sin 4 cos 3 θ = θ Let, sinθ = 4K and cosθ = 3K ∴ 9sin 12cos 36K 36K 27cos 20sin 81K 80K θ+ θ + = θ− θ − = 72 33. In a triangle ABC, right-angled at C, if 13 secA = , 5 then find the value of 1 + sinA cosB ? (a) 25 12 (b) 3 2 (c) 18 5 (d) 5 SSC CHSL 06/08/2021 (Shift-I) Ans. (a) : Q sec A = 13 5 Q BC 2 = 13 2 –5 2 BC = 12 ∴ 12 1 1 sin A 13 12 cos B 13 + + = = 25 12 34. If 5 cosecθ = 2 , then what will be the value of (secθ + tanθ – cotθ sinθ)? (a) 2 5 + (b) 2 5 2 5 + (c) 4 5 2 5 + (d) 5 2 5 + SSC CHSL 12/04/2021 (Shift-I) Ans : (c) 5 cosecθ = 2 secθ + tanθ – cotθ sinθ 5 2 1 2 1 1 2 5 + − × 5 2 5 1 5 + − = 4 2 5 + = 4 5 2 5 + 35. If (sin A – cos A) = 0, then what is the value of cot A? (a) 6 π (b) 0 (c) 1 (d) 4 π SSC CHSL 04/08/2021 (Shift-I) Ans. (c) : sin A – cos A = 0 sinA = cosA sin A 1 cos A = tan A = 1 cot A = 1 tan A (tan A = 1) cot A = 1 1 cot A = 1 Trick:– sinA–cosA= 0 sinA = cosA taking ∠A = 45º sin45º = cos45º 1 2 = So, cot45º= 1

Trigonometry 109 . 36. If 2 4x cosθ = 1+4x , then what is the value of sin θ? (a) 2 2 1 4x 1 4x + − (b) 2 2 1 4x 4x + (c) 2 2 1 4x 1 4x − + (d) 2 2 1 4x 4x − SSC CHSL 04/08/2021 (Shift-I) Ans. (c) : cosθ = 2 4x 1 4x + cosθ = B H H 2 = P 2 + B 2 (1+4x 2 ) 2 = P 2 + 16x 2 1+16x 4 + 2×1×4x 2 = P 2 + 16x 2 P 2 = (1+16x 4 –8x 2 ) P = (1–4x 2 ) sin θ = P H sin θ = 2 2 1 4x 1 4x − + 37. If 5 sin 2 θ = 3(1+cosθ), 0° < θ < 90°, then the value of cosecθ + cotθ is: (a) 4 21 (b) 3 7 (c) 5 21 (d) 7 3 SSC CHSL 13/04/2021 (Shift-I) Ans. (d) : 5sin 2 θ = 3 (1+cosθ) 5(1–cos 2 θ) = 3(1+cosθ) 5(1+cosθ) (1–cosθ) = 3 (1+cosθ) 5–5 cosθ = 3 cosθ = 2 5 = B H P = 25 4 21 − = cosec θ + cot θ = 5 2 7 7 3 21 21 21 + = = 38. If cotθ = 2 + 1, then cosecθ secθ = ? (a) 2 2 (b) 2 4 (c) 2 2 (d) 4 2 SSC CHSL 13/04/2021 (Shift-I) Ans. (c) : cot θ = 2 1 B 1 P + = H = ( ) () 2 2 2 1 1 + + = ( ) 3 2 2 1 4 2 2 + + = + cosec θ . sec θ = 4 2 2 4 2 2 1 2 1 + + × + = ( ) ( ) 2 2 2 1 4 2 2 2 2 2 1 2 1 + + = = + + 39. If 3sin 2 A + 4cos 2 A – 3 = 0, then the value of cot A (where 0 ≤ A ≤ 90º) is: (a) ∞ (b) 0 (c) 1 (d) not defined SSC CHSL 11/08/2021 (Shift-I) Ans. (b) : 3sin 2 A + 4cos 2 A – 3 = 0 ∴ given 0 ≤ A ≤ 90º Let ∠A = 90° Then, 3sin 2 A + 4cos 2 A – 3 = 3sin 2 90°+4cos 2 90°–3 Q L.H.S. = 3×1+0–3 = 0 = R.H.S. So, cot A = cot 90° = 0 40. If 41 cosec θ = 9 and θ is an acute angle, then the value of 5 tan θ will be: (a) 7 8 (b) 9 8 (c) 13 4 (d) 11 8 SSC CHSL 04/08/2021 (Shift-II) Ans. (b) : 41 hypotenuse cosec 9 perpendicular θ= = Q ∴ from triplet, base = 40 (Triplet, 41, 40, 9) Then 9 9 5 tan 5 40 8 θ= × = Perpendicular tan Base   θ=     Q 41. If 7 cosθ = 3 6 and θ is an acute angle, then the value of 2 3 27sin θ - 2 is: (a) 12 (b) 15 (c) 1 (d) 9 SSC CHSL 05/08/2021 (Shift-III) Ans. (c) : 7 cos 3 6 θ= 2 2 2 7 49 5 sin 1 cos 1 1 54 54 36   θ= − θ= − = − =     Q 2 3 5 3 5 3 2 27 sin 27 1 2 54 2 2 2 2 ∴ θ− = × − = − = =

Trigonometry 110 . 42. If 21 tanθ = 20, then (1+sinθ – cosθ): (1 – sinθ + cosθ) is equal to: (a) 11 : 13 (b) 14 : 15 (c) 12 : 11 (d) 13 : 15 SSC CHSL 16/04/2021 (Shift-III) Ans.(b) : 21tanθ = 20 (Given) 20 tan 21 θ= ( )( ) 1 sin cos :1 sin cos ∴ + θ− θ − θ+ θ 20 21 20 21 1 :1 29 29 29 29    = + − − +       28 30 : 28 : 30 14 :15 29 29 = = = 43. If sec θ = 65 63 and θ is an acute angle, then the value of 8 (cosecθ – cotθ) is: (a) 2 (b) 1 (c) 8 (d) 4 SSC CHSL 12/04/2021 (Shift-III) Ans : (b) Given, 65 sec 63 θ= 16 sin 65 θ= ( ) ( ) 63 8 1 1 cos 65 8 cosec cot 8 sin 16 65   × −   − θ   ∴ θ− θ= = θ 16 1 16 = = 44. If 2 2 P -1 cosθ = , 0º < θ < 90º , P +1 then cosec θ is equal to: (a) 2 1 P 2P + (b) 2 1 P 2P − (c) 2 2P 1 P − (d) 2 2P 1 P + SSC CHSL 05/08/2021 (Shift-II) Ans. (a) : 2 2 P 1 cos P 1 − θ= + ∵ sin 2 θ= 1– cos 2 θ ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 P 1 P 1 P 1 sin 1 P 1 P 1 + − −   − ∴ θ= − =   +   + ( ) 2 2 2 4P P 1 = + ( ) 2 2P sin P 1 θ= + 2 1 P cosec 2P + ∴ θ= 45. If 3cos 2 θ – 4sinθ + 1 = 0, 0°< θ < 90°, then the value of 3cos 2 θ + 5tan 2 θ will be: (a) 2 6 3 (b) 1 5 5 (c) 2 5 3 (d) 4 5 5 SSC CHSL 19/08/2021 (Shift-II) Ans. (c) : 3cos 2 θ – 4sinθ + 1 = 0 3(1–sin 2 θ) –4sinθ+1 = 0 3–3sin 2 θ – 4sinθ +1 = 0 3sin 2 θ + 4sinθ – 4 = 0 (sinθ + 2) (3sinθ–2) = 0 ( ) 2 sin ,sin 2 invalid 3 θ= θ=− 2 4 5 cos 1 sin 1 9 3 θ= − θ= − = Q ( ) ( ) 2 2 2 2 23 5 3cos 5 tan 3 5 9 53 θ+ θ= × + × 5 4 5 3 5 = + × 17 2 5 3 3 = = 46. If 5 cos θ = 4 sin θ, 0º ≤ θ ≤ 90º, then what will be the value of sec θ? (a) 41 5 (b) 3 5 (c) 41 16 (d) 41 4 SSC CHSL 09/082021 (Shift-II) Ans. (d) : 5cosθ = 4sinθ 5 tan 4 ∴ θ= ∵sec 2 θ = 1 + tan 2 θ

Trigonometry 111 . 2 25 41 sec 1 16 16 ∴ θ= + = 41 sec 4 θ= 47. If 5tanA = 12, then what is the value of 13sinA + 20tanA 15tanA - 13cosA , where A is an acute angle? (a) 19 1 41 (b) 31 41 (c) 41 50 (d) 29 1 31 SSC CHSL 12/08/2021 (Shift-II) Ans. (d) : 5tanA = 12 12 tan A 5 = 12 12 13 20 13sin A 20 tan A 13 5 5 12 15 tan A 13cosA 15 13 5 13 × + × + ∴ = − × − × 12 48 60 29 1 36 5 31 31 + = = = − 48. If cosec A = 10, then what is the value of 20 sinA + 9 11 sec A? Given that is an acute angle. (a) 32 (b) 34 (c) 30 (d) 23 SSC CHSL 12/08/2021 (Shift-II) Ans. (a) : cosesA = 10 or, 2 2 1 sin A cos A 1 sin A 10 = = − Q 1 99 cos A 1 100 10 ∴ = − = 1 10 20 sin A 9 11secA 20 9 11 10 99 ∴ + = × + × =2 +30 = 32 49. If tan θ = 15, then what is the value of sec θ ? (a) 224 (b) 226 (c) 1 226 (d) 1 224 SSC CHSL 15/04/2021 (Shift-II) Ans : (b) tanθ = 15 ∵sec 2 θ = 1 + tan 2 θ ∴ sec 2 θ = 1 + (15) 2 = 1 + 225 = 226 sec 226 θ= 50. If tan A = 1.1 6 , then what is the value of (4 cos A – 7 sinA)? Given that A is an acute angle. (a) 2 14 61 (b) 2 41 71 (c) 2 41 61 (d) 2 14 71 SSC CHSL 13/04/2021 (Shift-II) Ans. (c) : 1.1 11 tan A 6 60 = = ( ) 60 11 4 cos A 7 sin A 4 7 61 61 − = × − × 240 77 41 2 61 61 −   = =     51. If 9 sinB = , 41 then what is the value of cot B, where 0º < B < 90º? (a) 41 9 (b) 40 9 (c) 9 41 (d) 9 40 SSC CHSL 09/08/2021 (Shift-III) Ans. (b) : 9 sin B 41 = 2 cos B 1 sin B = − Q 81 1600 40 cos B 1 1681 1681 41 ∴ = − = = cos B 40 41 40 cot B 9 sin B 9 41 = = = Q 52. If sinθ = 2 ab , a > b > o, a+b then the value of 1 θ cos + cosθ - 1 will be: (a) a b − (b) a b (c) b a − (d) b a SSC CHSL 19/08/2021 (Shift-II) Ans. (a) : 2 ab sin a b θ= + 2 cos 1 sin θ= − θ Q ( ) ( ) ( ) 2 2 2 a b 4ab 4ab a b cos 1 a b a b a b + − − ∴ θ= − = = + + + cos 1 a b a b 2a a cos 1 a b a b 2b b θ+ − + + − ∴ = = = θ− − − − −

Trigonometry 112 . 53. If 3 sin 2 θ – cosθ – 1 = 0, 0 0 < θ < 90 0 , then what is the value of cot θ + cosecθ ? (a) 3 2 2 (b) 2 3 (c) 5 (d) 2 5 SSC CGL–(Tier-I) 18/08/2021 (Shift I) Ans. (c) : 3 sin 2 θ – cosθ – 1 = 0 0 0 < θ < 90 0 3 (1–cos 2 θ) – cos θ –1 = 0 3 – 3 cos 2 θ – cos θ–1 = 0 3 cos 2 θ + cosθ – 2 = 0 3cos 2 θ + 3 cosθ – 2cosθ – 2 = 0 3 cos θ (cosθ +1) – 2 (cosθ +1) = 0 (cosθ + 1) (3cosθ–2) = 0 cosθ = –1 (invalid) and 2 B cos 3 H θ= = H 2 = P 2 + B 2 P 2 = 3 2 – 2 2 P = 5 cot θ + cosecθ = ? 2 3 5 5 5 5 5 + = = 54. Find the value of sin(60+θ) – cos(30–θ). (a) 1 (b) 1 2 (c) 0 (d) -1 SSC CHSL –18/03/2020 (Shift-II) Ans. (c) : sin(60+θ) – cos(30–θ) sin(60+θ) – sin(60+θ) = 0 55. If o o x sin 30 cos 60 = sin 45° cos 45°, then the value of x is: (a) 2 (b) 0 (c) 3 (d) 1 SSC CHSL –19/03/2020 (Shift-III) Ans. (a) : o o x sin 30 cos 60 = sin 45° cos 45° 1 1 1 1 x 2 2 2 2 × × = × x 1 4 2 = 4 x 2 = ⇒ Hence x = 2 56. If A lies in the first quadrant and 6 tanA = 5, then the value of 8sinA - 4cosA cosA + 2sinA is: (a) 4 (b) 16 (c) –2 (d) 1 SSC CGL (Tier-I)-2019 – 03/03/2020 (Shift-I) Ans. (d) : 6tanA = 5 5 tan A 6 = 8sinA 4cosA cos A 2sinA − ∴ + On dividing by cosA in each term, 5 8 4 8 tan A 4 6 5 1 2 tan A 1 2 6 × − − = = + + × 16 1 16 = = 57. If secθ + tanθ = p, 0° < θ < 90°, then 2 2 p -1 p +1 is equal to: (a) 2cosecθ (b) sinθ (c) cosecθ (d) cosθ SSC CGL (Tier-I) – 04/03/2020 (Shift-III) Ans. (b) : secθ + tanθ = p.......... (1) ( ) 2 2 1 sec tan sec tan 1 .........(2) p θ− θ= θ− θ= Q On adding, 1 2sec p p θ= + , 2 p 1 sec 2p + θ= 2 2 p 1 sin p 1 − ∴ = θ + 58. If secθ – tanθ = x y , (0 < x < y) and 0° < θ < 90°, then sinθ is equal to: (a) 2 2 2 2 x y y x + − (b) 2 2 2xy x y + (c) 2 2 2 2 y x x y − + (d) 2 2 x y 2xy + SSC CGL (Tier-I) – 04/03/2020 (Shift-I) Ans. (c) : x sec tan y θ− θ=

Trigonometry 113 . 2 2 y sec tan sec tan 1 x   θ+ θ= θ− θ=   Q 2 2 x y x y 2sec y x xy + ∴ θ= + = 2 2 x y sec 2xy + θ= ( ) 2 2 2 2 2 2 2 AB x y 4x y y x = + − = − [ ] y x > Q 2 2 2 2 y x sin x y − ∴ θ= + 59. If 5cosθ – 12sinθ = 0, then what is the value of 1 + sinθ + cosθ 1 - sinθ + cosθ ? (a) 5 4 (b) 3 2 (c) 5 2 (d) 3 4 SSC CGL (Tier-I)– 05/03/2020 (Shift-II) Ans. (b) : 5 cosθ – 12 sinθ = 0 cotθ = 12 5 1 sin cos 1 sin cos + θ+ θ − θ+ θ = 5 12 1 13 13 5 12 1 13 13 + + − + = 30 3 13 20 2 13 = 60. If 0º < A, B < 45°, cos(A+B) = 24 25 and sin(A–B) = 15 17 , then tan2A is: (a) 416 87 (b) 0 (c) 1 (d) 213 4 SSC CGL (Tier-I) – 06/03/2020 (Shift-III) Ans. (a) : tan2A = tan[(A+B) + (A–B)] = tan(A B) tan(A B) 1 tan(A B).tan(A B) + + − − + − = 7 15 24 8 7 15 1 24 8 + − × = 52 52 64 416 24 29 24 29 87 64 × = = × 61. If sinA + cosA 17 = cosA 12 , then the value of 1 - cosA sinA is: (a) 5 12 (b) –5 (c) 1 (d) 1 5 SSC CGL (Tier-I) – 07/03/2020 (Shift-II) Ans. (d) : sin A cos A 17 cos A 12 + = 17 tan A 1 12 + = 5 tan A 12 = 12 1 1 cos A 13 5 sin A 13 − − ∴ = 1 5 = 62. The value of 2 2 4 4 1- 2sin θcos θ -1 sin θ + cos θ is: (a) –1 (b) 1 (c) -2sin 2 θcos 2 θ (d) 0 SSC CGL (Tier-I)– 07/03/2020 (Shift-I) Ans. (d) : 2 2 4 4 1 2sin cos 1 sin cos − θ θ − θ+ θ Q (sin 2 θ + cos 2 θ) 2 = sin 4 θ + cos 4 θ + 2sin 2 θ.cos 2 θ ∴sin 4 θ + cos 4 θ = 1–2sin 2 θ.cos 2 θ then, 4 4 4 4 sin θ + cos θ -1 sin θ + cos θ 1 – 1 = 0

Trigonometry 114 . 63. If sec θ (cosθ + sinθ) = 2 , then what is the value of (2 sin θ)/(cos θ – sinθ) ? (a) 3 2 (b) 3/ 2 (c) 1/ 2 (d) 2 SSC CGL (Tier-II) 20-02-2018 Ans. (d) : ( ) sec cos sin 2 θ θ+ θ= 1 tan 2 + θ= tan 2 1 θ= − According to the question, 2sin 2 tan cos sin 1 tan θ θ = θ− θ − θ ( ) ( ) 2 2 1 1 2 1 − = − − ( ) ( ) 2 2 1 2 2 1 − = − 2 = 64. If 2 2 1 + sin p = 1-sin q φ φ , then secφ is equal to : (a) 2 2 1 1 p q + (b) 2 2 2 2 pq p q + (c) 1 q p 2 p q   +     (d) 2 2 2 2 2p q p q + SSC CGL (Tier-II) 12-09-2019 Ans. (c) 2 2 1 sin p 1 sin q + φ = − φ On multiplying by (1 + sinφ) in numerator and denominator ( ) 2 2 2 2 2 2 2 2 1 sin p 1 sin 2sin p 1 sin q cos q + φ + φ+ φ = ⇒ = − φ φ 2 2 2 2 p sec tan 2sec tan q φ+ φ+ φ φ= ( ) 2 2 2 p sec tan q ∴ φ+ φ = p sec tan ........(i) q φ+ φ= q sec tan ........(ii) p ∴ φ− φ= {Q sec 2 φ – tan 2 φ = 1} By adding the equation (i) and (ii) p q 2sec q p   φ= +     1 p q sec 2 q p   φ= +     65. If ( ) 2 cosθ = 2p/ 1 + p , then tanθ is equal to: (a) 2 2 p 1 p + (b) 2 2 1 p 1 p − + (c) 2 2p 1 p − (d) 2 1 p 2p − SSC CGL (TIER-I) – 04.06.2019 (Shift-II) Ans. (d) Q 2 2p cos 1 p θ= + ∴ 2 1 p tan 2p − θ= 66. If 3sinθ = 2cosθ, then 4sin cos 4cos sin θ− θ θ− θ θ− θ θ− θ θ+ θ θ+ θ θ+ θ θ+ θ is equal to: (a) 5 11 (b) 5 14 (c) 5 7 (d) 5 8 SSC CGL (TIER-I) – 13.06.2019 (Shift-I) Ans. (b) : According to the question, 3sin 2cos θ= θ sin 2 cos 3 θ ⇒ = θ 2 tan 3 θ= 4sin cos 4 cos sin θ− θ ∴ θ+ θ 4 tan 1 4 tan θ− ⇒ + θ [On dividing by cos θ in numerator and denominator] 2 4 1 3 2 4 3 × − ⇒ + 8 3 5 3 12 2 14 3 −       ⇒ = +       67. If cosec θ = 1.25, then 4tanθ - 5cosθ + 1 secθ + 4cotθ - 1 = ? (a) 2 (b) 9 10 (c) 1 2 (d) 10 11 SSC CPO-SI – 12/12/2019 (Shift-I)

Trigonometry 115 . Ans. (d) cosecθ = 1.25 = 5 4 ∴ 4 3 4 5 1 4 tan 5cos 1 3 5 5 3 sec 4 cot 1 4 1 3 4 × − × + θ− θ+ = θ+ θ− + × − 16 3 1 3 5 3 1 3 − + = + − 10 3 11 3 = 10 11 = 68. If 2sinθ + 15cos 2 θ = 7, 0º < θ < 90º, then what is the value of 3 - tanθ 2 + tanθ ? (a) 1 2 (b) 5 8 (c) 1 4 (d) 3 4 SSC CPO-SI – 12/12/2019 (Shift-I) Ans. (a) 2sinθ + 15cos 2 θ = 7 2sinθ + 15(1 – sin 2 θ) = 7 15sin 2 θ – 2sinθ – 8 = 0 15sin 2 θ – 12sinθ + 10sinθ – 8 = 0 (5sinθ – 4) (3sinθ + 2) = 0 Sinθ = 4 5 or 2 3 − ∴ tanθ = 4 3 Now, 4 3 3 tan 3 4 2 tan 2 3 − − θ = + θ + ⇒ 5 1 3 10 2 3       =       69. If 21 tanθ = 20, then (1 + sinθ + cosθ) : (1 – sinθ + cosθ) = ? (a) 7 : 3 (b) 5 : 2 (c) 3 : 1 (d) 2 : 1 SSC CPO-SI – 11/12/2019 (Shift-II) Ans. (a) 21 tanθ = 20 tanθ = 20 21 then, = (1 + sinθ + cosθ) : (1 – sinθ + cosθ ) = 20 21 20 21 1 :1 29 29 29 29    + + − +       = 70 30 : 7:3 29 29 = 70. If (cosθ + sinθ) : (cosθ – sinθ) = ( 3 + 1 ) : ( 3 – 1), 0 0 < θ < 90 0 then what is the value of secθ? (a) 2 (b) 2 3 3 (c) 2 (d) 1 SSC CPO-SI – 09/12/2019 (Shift-II) Ans. (b) (cosθ + sinθ) : (cosθ – sinθ) = ( 3 + 1) : ( 3 – 1) cos sin cos - sin θ+ θ θ θ = 3 1 3 1 + − From componendo and dividendo Rule, cos 3 sin θ = θ or 1 tan 3 θ= tan 2 θ = 1 3 sec 2 θ –1 = 1 3 sec 2 θ = 1 1 3 + = 4 3 secθ = 2 2 3 3 3 = 71. The value of             3 3 3 + 2sin P 1 + 2cos P + 1 – 2cos P 3 – 2sin P is: (a) sinP cosP (b) 2sinP cosP (c) 1 (d) 0 SSC CHSL –13/10/2020 (Shift-I) Ans. (d) : 3 3 3 + 2sin P 1 + 2cos P 1 – 2cos P 3 2sin P     +     −     Let P = 90 0 3 3 0 0 0 0 3 2sin 90 1 2cos 90 1 2cos 90 3 2sin 90     + + +     − −         3 3 3 2 1 1 0 1 0 3 2×1   + × +   +     − −    

Trigonometry 116 . ( ) 3 3 1 3 2 3 2   + +   −   ( ) ( ) ( ) 3 3 3 2 3 2 3 2 3 2   +   + +   − +   ( ) 3 3 3 2 3 2 3 4   + + +   −   ( ) ( ) 3 3 3 2 3 2 + − + = 0 72. If ( ) 2 2P cos θ ,p 1 P 1 = ≠± + then cosecθ is equal to? (a) 2 2 P 1 P 1 + − (b) 2 2P P 1 + (c) 2 P 1 2P − (d) 2 2P P 1 − SSC CHSL (Tier-I) 05/07/2019 (Shift-I) Ans. (a) : 2 2P cos θ P 1 = + ( ) 2 2 2 2 2P 1 sin θ cos θ 1 sin θ P 1 − = = − + 2 2 2 2P 1 sin θ P 1   − =   +   s 2 2 2 2 4P sin θ 1 (P 1) = − + 2 2 2 2 2 2 (P 1) 4P sin θ (P 1) + − = + 4 2 2 2 2 2 P 1 2P 4P sin θ (P 1) + + − = + 4 2 2 2 2 P 1 2P sin θ (P 1) + − = + 2 2 2 2 2 (P 1) sin θ (P 1) − = + 2 2 P 1 sin θ P 1 − = + 2 2 P 1 cosec θ P 1 + = − 73. Find x if 2 sin 2 x – 1 = 0. (a) 4 π (b) π (c) 0 (d) 2 π SSC CHSL –17/03/2020 (Shift-III) Ans. (a) : 2 sin 2 x – 1 = 0 2 sin 2 x = 1 sin 2 x 1 2 = 1 sin x sin 45º 4 4 2 π π   = = =     ⇒ x 4 π = 74. If sinθ = 4 5 , find the value of sin 3θ. (a) 64 125 (b) 32 45 (c) 12 25 (d) 44 125 SSC CHSL –17/03/2020 (Shift-III) Ans. (d) 4 sin 5 θ= ∴ sin3θ = 3sinθ – 4sin 3 θ 3 4 4 3 –4 5 5   = × ×     12 4 64 – 5 125 × = 12 25 – 4 64 125 × × = 300 – 256 125 = 44 125 = 75. If sec A = 5 3 , then what is the value of cot A? (a) 3 4 (b) 3 5 (c) 4 5 (d) 4 3 SSC CHSL –19/10/2020 (Shift-III) Ans. (a) : sec A = 5 3 Q sec 2 A – 1 = tan 2 A 2 25 1 tan A 9 − = , tan 2 A = 16 9 cot 2 A = 9 16 ⇒ cot A = 3 4 76. If 0 < θ < 90 o , 3b cosecθ = a secθ and 3a secθ – b cosecθ=8 then the value of 9b 2 + a 2 is: (a) 6 (b) 8 (c) 7 (d) 9 SSC CHSL –19/10/2020 (Shift-II)

Trigonometry 117 . Ans. (d): 3b cosecθ = a secθ ____(i) or 9b cosecθ = 3a secθ Q 3a secθ – b cosecθ = 8 ⇒ 9b cosecθ – b cosecθ = 8 b cosecθ = 1 b sin = θ Form equation (i) 3×sinθ cosecθ = a sec θ a 3cos = θ ∴ 9b 2 + a 2 = 9 sin 2 θ + 9 cos 2 θ = 9 77. If tanθ + cotθ = 6, then find the value of tan 2 θ + cot 2 θ. (a) 26 (b) 24 (c) 36 (d) 34 SSC CHSL –15/10/2020 (Shift-II) Ans. (d) : tanθ + cotθ = 6 On squaring both sides, (tanθ + cotθ) 2 = 6 2 tan 2 θ + cot 2 θ + 2tanθcotθ = 36 tan 2 θ + cot 2 θ = 34 78. If sinx = 2 3 , then find the value of cos 3x. (a) 0.5678 (b) -0.8765 (c) 0.6735 (d) -0.5797 SSC CHSL –15/10/2020 (Shift-II) Ans. (d) : sinx = 2 2 cos x 1 sin x 3 ⇒ = − = 4 1 9 − = 5 3 Q cos 3x = 4cos 3 x – 3cosx, = 4 55 3 5 27 3 × × − = 20 5 27 5 27 − = 7 5 0.5797 27 − =− 79. If tan x = 3 2 , then the value of 3sinx + 2cosx 3sinx – 2cosx is: (a) 13 5 (b) 5 (c) 5 13 (d) 1 5 SSC CHSL –14/10/2020 (Shift-I) Ans. (a) : tan x = 3 2 3sin x 2cosx 3sin x – 2 cos x + 3 3 2 3tan x 2 13 2 13 2 3 tan x 2 5 2 5 × + + = = − × 80. If cosx = ≤ ≤ º 24 ,0º x 90 25 , then the value of cotx + cosecx is: (a) 7 (b) 0 (c) 7 2 (d) 1 SSC CHSL –14/10/2020 (Shift-I) Ans. (a) : cosx = 24 25 AB = 625 576 49 7 − = = cot x + cosec x = 24 25 7 7 7 + = 81. If A is an acute angle and cot A + cosec A = 3, then the value of cos A is equal to: (a) 4 5 (b) 2 5 (c) 1 5 (d) 3 5 SSC CHSL –26/10/2020 (Shift-I) Ans. (a) : cotA + cosecA = 3 ____(i) Q cosec 2 A – cot 2 A = 1 (cosecA – cotA) (cosecA + cotA) = 1 cosecA – cotA = 1 3 ____(ii) On adding the equation (i) and (ii). cotA + cosecA = 3 cosecA – cotA = 1 3 1 2cosecA 3 3 = + cosec A = 5 3 From Triplet, Base = 4 Base cos A Hypotenuse ∴ = 4 5 =

Trigonometry 118 . 82. If cosec θ = 13 12 , then the value of 2sinθ – 3cosθ 4sinθ – 9cosθ is: (a) 2 (b) 4 (c) 2 (d) 3 SSC CHSL –26/10/2020 (Shift-I) Ans. (d) : Given, cosecθ = 13 12 (AB) 2 = (AC) 2 + (BC) 2 (13) 2 = (12) 2 + (BC) 2 169 = 144 + (BC) 2 (BC) 2 = 169 – 144 = 25 BC = 5 ∴ 12 5 2 3 2sin 3cos 13 13 12 5 4sin 9cos 4 9 13 13 × − × θ− θ = θ− θ × − × 24 15 9 13 13 13 48 45 3 13 13 13 − = − = 3 83. If θ 20 tan = 21 , then the value of θ θ θ θ sin – cos sin + cos is: (a) –29 31 (b) 27 21 (c) 29 35 (d) 1 41 − SSC CHSL–12/10/2020 (Shift-III) Ans. (d) : 20 tan = 21 θ ⇒ sin cos sin + cos – θ θ θ θ 20 –1 tan –1 –1 21 20 tan 1 41 1 23 θ = = θ+ + 84. If 2 cotθ = 3, then 13 cosθ - 3 tanθ 3 tanθ + 13 sinθ is: (a) 2 3 (b) 1 5 (c) 3 4 (d) 1 4 SSC CHSL –19/03/2020 (Shift-II) Ans. (d) Q 2 cot θ = 3 or cot θ = 3 2 or tan θ = 2 3 QHypotenuse 2 = Base 2 + Height 2 (AC) 2 = 3 2 + 2 2 (AC) 2 = 9 + 4 (AC) 2 = 13 AC = 13 ∴ 13 cos 3tan 3tan 13 sin θ− θ θ+ θ = 3 2 13. 3. 3 13 2 2 3. 13. 3 13 − + = 3 2 1 2 2 4 − = + 85. If sinx + cosx 6 = sinx - cosx 5 , then the value of 2 2 tan x + 1 tan x-1 is: (a) 61 60 (b) 60 61 (c) 35 61 (d) 61 35 SSC CHSL –19/03/2020 (Shift-I) Ans. (a) : sin x cos x 6 sin x cos x 5 + = − or 5 sin x + 5 cos x = 6 sin x – 6 cos x or 11 cos x = sin x or sin x 11 cos x = or tan x = 11 ∴ ( ) ( ) 2 2 2 2 11 1 tan x 1 tan x 1 11 1 + + = − − = 121 1 121 1 + − = 122 120 ⇒ 61 60 86. If secθ + tanθ =5 secθ - tanθ and θ is an acute angle, then the value of 2 2 3cos θ + 1 3cos θ-1 is: (a) 1 (b) 4 (c) 3 (d) 2 SSC CHSL –21/10/2020 (Shift-II)

Trigonometry 119 . Ans. (b) sesθ + tan 5 sec - tan θ = θ θ From componendo and dividendo rule secθ + tan sec tan 5 1 sec + tan sec tan 5 1 θ+ θ− θ + = θ θ− θ+ θ − 2sec 6 2 tan 4 θ = θ 1 2 cos sin 3 cos θ = θ θ cosecθ = 3 2 AC 2 = AB 2 + BC 2 (3) 2 = (2) 2 + (BC) 2 9 = 4 + (BC) 2 9 4 BC − = BC = 5 2 2 3cos 1 then, 3cos 1 θ+ θ− 2 2 5 3 1 3 5 3 1 3   × +         × −       = 5 3 1 9 5 3 1 9 × + × − 5 8 1 3 3 5 2 1 3 3 + = − = 4 87. If secθ + tanθ = 2 + 5 and θ is an acute angle, then the value of sin θ is: (a) 3 5 (b) 2 5 5 (c) 1 5 (d) 5 5 SSC CHSL –21/10/2020 (Shift-III) Ans. (b) Given, secθ + tanθ = 2 + 5 ...(i) Then sinθ = ? Q sec 2 θ – tan 2 θ = 1 (secθ + tanθ) (secθ – tanθ) = 1 (secθ – tanθ) = 1 2 5 2 5 2 5 − × + − = 2 5 5 2 1 − = − − ...(ii) On adding the equation (i) and (ii), secθ + tanθ = 2 + 5 secθ – tanθ = 5 – 2 2sec θ = 2 +2 5 – 2 sec θ = 5 cos θ = 1 5 sinθ = 2 5 5 5 × , sinθ = 2 5 5 88. If 1 cot 3 θ= θ= θ= θ= then what is the value of ( ) 2 2 2 2 sin cosec sec 1 cos − θ − θ − θ − θ + θ+ θ + θ+ θ + θ+ θ + θ+ θ − θ − θ − θ − θ (a) 7 (b) 4 (c) 6 (d) 5 SSC CHSL 02/07/2019 (Shift-I) Ans. (d) : 1 cot 3 θ= then ( ) 2 2 2 2 sin cosec sec 1 cos − θ + θ+ θ − θ 3 2 4 4 2 1 3 1 4 −   = + +     − 5 4 10 4 3 3 = × + 15 5 3 = = (II) Problems based on Trigonometric Identities 89. The given expression (cot B – tan B). sinB.cosB is equal to : (a) 2cos 2 B – 1 (b) 2sec 2 B – 1 (c) 1 – 2sec 2 B (d) 1 – 2cos 2 B SSC CHSL (Tier-I) 14/08/2023 (Shift-IV) Ans. (a) : Given that, (cot B – tan B). sin B cos B ⇒ cos B sin B sin B cos B   −     . sin B cos B ⇒ 2 2 cos B sin B sin B.cos B   −     . sinB cos B = cos 2 B – sin 2 B = cos 2 B – (1 – cos 2 B) = 2cos 2 B – 1

Trigonometry 120 . 90. cos θ cos θ + sec θ – 1 sec θ + 1 is equal to : (a) 2 sin 2 θ (b) 2 cot 2 θ (c) 2 cos 2 θ (d) 2 sec 2 θ SSC CHSL (Tier-I) 02/08/2023 (Shift-I) Ans. (b) : cos cos sec 1 sec 1 θ θ + θ− θ+ 1 1 cos sec 1 sec 1   = θ +   θ− θ+   ( )( ) sec 1 sec 1 cos sec 1 sec 1   θ+ + θ− = θ   θ− θ+   2 2 2 cos cos 2sec cos sec 1 tan θ× θ× θ θ = = θ− θ 2 2 1 sec cos sec 1 tan   θ=   θ   θ− = θ     Q 2 2 cot = θ 91. If cotθ = 9 17 then find the value of cosec 2 θ. (a) 8 17 (b) 9 1 17 (c) 81 1 289 (d) 208 289 SSC CHSL (Tier-I) 02/08/2023 (Shift-I) Ans. (c) : 9 cot 17 θ= 2 2 81 81 cos ec 1 cot 1 1 289 289 θ= + θ= + = Q 92. If tanθ + secθ = a, then secθ is equal to? (a) 2 a 1 2a + (b) 2 a 1 2a + (c) 2 (a 1) 2a − (d) 2 a 1 2a − SSC CHSL (Tier-I) 08/08/2023 (Shift-II) Ans. (a) : tan sec a...........(i) θ+ θ= On squaring both the sides, 2 2 2 tan sec 2 tan sec a θ+ θ+ θ θ= 2 2 2 sec 1 sec 2 tan sec a θ− + θ+ θ θ= 2 2 [ sec 1 tan ] θ− = θ Q 2 2 2sec 2 tan sec a 1 θ+ θ θ= + 2 2sec (sec tan ) a 1 θ θ+ θ= + 2 2sec a a 1 θ× = + (By equation (i)) 2 a 1 sec 2a + θ= 93. If sinA = 3 5 then calculate the value of cosA + tanA –1 : (a) 2 (b) 11 20 (c) 21 20 (d) 13 20 SSC CHSL (Tier-I) 10/08/2023 (Shift-I) Ans. (b) : 3 P sin A 5 H = = By Pythagoras theorem in ΔABC, AB 25 9 4 = − = 4 3 cos A tan A 5 4 = = 4 3 cos A tan A 1 1 5 4 + − = + − 16 15 20 11 20 20 + − = = 94. If sinθ + cosecθ = 5 then the value of (sin 3 θ + cosec 3 θ) will be : (a) 1 5 (b) 2 5 (c) 0 (d) 3 5 SSC CGL (Tier-I) 24/07/2023 (Shift-III) Ans. (b): sin cosec 5 θ+ θ= then 3 3 sin cos ec ? θ+ θ= Formula 3 3 3 a b (a b) 3ab(a b) + = + − + 3 3 3 sin cos ec (sin cosec ) 3sin .cosec (sin cosec ) θ+ θ= θ+ θ − θ θ θ+ θ ( ) ( ) 3 1 5 3sin . 5 sin = − θ θ 55 35 2 5 = − = 95. If tanθ + secθ = a, then secθ is ? (a) 1 2a (b) 2 a 1 2a + (c) a 1 2a − (d) 2 a 1 2a − SSC Selection Posts XI-28/06/2023 (Shift-III) Ans. (b) : Given that tanθ + secθ = a..................(i) จ sec 2 θ – tan 2 θ = 1 ⇒ (secθ – tanθ) (secθ + tanθ) = 1 ง secθ – tanθ = 1 a .................(ii) By Equation (i) and (ii), 2secθ = 2 a 1 a + ⇒ secθ = 2 a 1 2a +

Trigonometry 121 . 96. If tanx = 7 5 , then find the value of 42 9sinx - cosx 5 15sinx + 21cosx . (a) 0.1 (b) 1 (c) 0 (d) 0.5 SSC CGL (Tier-I) 24/07/2023 (Shift-III) Ans. (a) : 7 tan x 5 = 9 sin x 42 42 cos x 9 sin x cos x cos x 5 5 15sin x 15sin x 21cos x cos x 21 cos x   − −     = +   +     42 7 42 9 tan x 9 21 5 5 5 5 7 15 tan x 21 42 15 21 5 − × − = = = + × + 1 0.1 10 = = 97. If 4tanA = 3 then (cos 2 A – sin 2 A) is equal to : (a) 3 5 (b) 1 5 (c) 7 25 (d) 9 25 SSC Selection Posts XI-27/06/2023 (Shift-III) Ans. (c) : 4 tan A = 3 tan A = 3 4 ∴ cos 2 A - sin 2 A = 2 2 4 3 5 5     −         16 9 25 25 = − 7 25 = 98. Get the value of       1 – sinθ sinθ . (a) cosθ secθ (b) cosθ tanθ (c) cosθ cosecθ (d) cosθ cotθ SSC CGL (Tier-I) 25/07/2023 (Shift-IV) Ans. (d) : 2 2 1 sin 1 – sin cos – sin 1 sin sin   θ θ θ   = =     θ θ θ     cos .cos cos .cot sin θ = θ= θ θ θ 99. Find the value of 1 – sin3 1 + sin3 θ θ . (a) sec 3θ - tan 3θ (b) sec 3θ + tan 3θ (c) (sec 3θ - tan 3θ) 3 (d) (sec 3θ - tan 3θ) 2 SSC CGL (Tier-I) 21/07/2023 (Shift-II) Ans. (a) : ( ) ( ) ( )( ) 1 sin 3 1 sin 3 1 sin 3 1 sin 3 1 sin 3 1 sin 3 − θ× − θ − θ = + θ + θ − θ ( ) 2 2 1 sin 3 1 sin 3 − θ = − θ ( ) 2 2 1 sin 3 1 sin 3 cos 3 cos3 − θ − θ = = θ θ 1 sin 3 cos3 cos3 θ = − θ θ sec 3 tan 3 = θ− θ 100. What will be the value of (secx – cosx)? (a) secx cosx (b) tanx cosx (c) secx tanx (d) tanx sinx SSC CGL (Tier-I) 21/07/2023 (Shift-II) Ans. (d): secx – cosx 2 1 1 cos x cos x cos x cos x − = − = 2 2 2 sin x [ 1 cos x sin x] cos x = − = Q sinx .sin x tan x sin x cos x = = 101. If 7cotP = 24 then find the value of sinP. (a) 7 25 (b) 625 7 (c) 49 625 (d) 24 25 SSC CGL (Tier-I) 27/07/2023 (Shift-III) Ans. (a) : 7 cot P = 24 ⇒ 24 B cot P 7 P = = In ∆PQR Pythagoras theorem 24 7 P Q R 2 2 PR 24 7 576 49 625 25 = + = + = = P 7 sin P H 25 = = 102. If tan 6 π + sec 6 π = x then find the value of x : (a) –1 3 (b) 1 3 (c) 3 (d) 2 3 SSC CGL (Tier-I) 17/07/2023 (Shift-II) Ans. (c) : tan sec x 6 6 π π + = tan30 0 + sec30 0 = x 1 2 3 x x x 3 3 3 3 + = ⇒ = ⇒ =

Trigonometry 122 . 103. If cosec θ + cot θ = 3 2 , then find the value of cosecθ. (a) 3 2 (b) 9 13 (c) 11 12 (d) 13 12 SSC CGL (Tier-I) 24/07/2023 (Shift-III) SSC CHSL (Tier-I) 11/08/2023 (Shift-I) Ans. (d): cosecθ + cotθ = 3 2 .........(i) On squaring both the sides, (cosecθ + cotθ) 2 = 9 4 cosec 2 θ + cosec 2 θ – 1 + 2 cosecθ cotθ = 9 4 (2 cosec 2 θ + 2 cosecθ cotθ) = 9 4 + 1 2 cosecθ (cosecθ + cotθ) = 13 4 2 cosecθ 3 13 2 4   × =     [By Equation. (i)] 13 13 cosec cosec 4 3 12 θ= ⇒ θ= × 104. Which of the following statement is incorrect? (a) cosec 2 A − 1 = cot 2 A (b) cos A × sec A = 1 (c) tan 2 A = 1 − sec 2 A (d) sin A = tan A × cos A SSC CGL (Tier-I) 18/07/2023 (Shift-III) Ans. (c) : 2 2 tan A 1 sec A = − This statement is not true because- 2 2 tan A sec A 1 = − 105. Select the incorrect formula from the given following options. (a) sec 2 θ+cos 2 θ = 1 (b) sec 2 θ–tan 2 θ = 1 (c) cosec 2 θ–cot 2 θ = 1 (d) sin 2 θ+cos 2 θ = 1 SSC CGL (Mains) 03/03/2023 Ans. (a) From the given options (a) sec 2 θ + cos 2 θ = 1(wrong) (b) sec 2 θ – tan 2 θ = 1 (Right) (c) cosec 2 θ – cot 2 θ = 1 (Right) (d) sin 2 θ + cos 2 θ = 1 (Right) Therefore, formula of option (a) is wrong. 106. If 8 Sin A = 17 then find the value of (cotA + secA). (a) 1 4 120 (b) 1 2 120 (c) 1 3 120 (d) 1 5 120 SSC CHSL Tier- II 26/06/2023 Ans.(c) : 8 sin A 17 = 2 cosA 1 sin A = − 2 8 1 17   = −     289 64 289 − = 225 289 = 15 cos A 17 = cos A 1 cot A sec A sin A cos A + = + 15/17 1 8/17 15/17 = + 15 17 8 15 = + 225 136 361 120 120 + = = 1 3 120 = 107. If sinx + sin 2 x = 1, then the value of cos 8 x + 2 cos 6 x+ cos 4 x is : (a) 1 (b) 0 (c) 2 (d) 4 SSC CHSL Tier- II 26/06/2023 Ans.(a) : sin x + sin 2 x = 1 sinx = 1–sin 2 x sinx = cos 2 x ........ (I) On squaring both sides (sinx + sin 2 x) 2 = (1) 2 sin 2 x + sin 4 x + 2sinx.sin 2 x = 1 sin 4 x + 2sin 3 x + sin 2 x = 1 ( ) ( ) ( ) 4 3 2 2 2 2 cos x 2 cos x cos x 1 + + = {From equation (i)} 8 6 4 cos x 2cos x cos x 1 + + = 108. If tanA.tanB + cos x =1 cosAcosB , then x = ? (a) B (b) A + B (c) A (d) A – B SSC CGL 08/12/2022 (Shift-IV) Ans. (b) : Given that, cosx tanA.tanB 1 cos A.cos B + = sin A.sin B cos x 1 cos A.cos B cos A.cos B + = sinA.sinB + cosx = cosA.cosB cosx = cosA.cosB–sinA.sinB cosx = cos(A+B) ง x = A+B

Trigonometry 123 . 109. Find the value of expression 2 2 2 2 sin 63° + sin 27° cos 17° + cos 73° . (a) 3 (b) 2 (c) 1 (d) 0 SSC CGL 08/12/2022 (Shift-IV) Ans. (c) : 2 2 2 2 sin 63º sin 27º cos 17º cos 73º + + 2 2 2 2 sin 63º sin (90 – 63º ) cos (90º –73º ) cos 73º + = + 2 2 2 2 sin 63º cos 63º 1 1 sin 73º cos 73º 1 + = = = + 110. If bcosθ = a, then cosec θ + cotθ will be? (a) b–a b+a (b) b+a b–a (c) 1 b–a (d) 1 b+a SSC CGL 09/12/2022 (Shift-IV) Ans. (b) : According to question, b cosθ = a cosθ = a b cosecθ + cotθ 2 2 2 2 b a b –a b –a + 2 2 b a b –a + b a b a b a b–a b a b–a + × + + = + × 111. What is the value of sin75º + sin15º? (a) 3 2 (b) 1 2 (c) 3 2 (d) 3 2 SSC CGL 09/12/2022 (Shift-IV) Ans. (a) : sin75° + sin15° By formula- sinC + sinD = ( ) ( ) C D C D 2sin .cos 2 2 + − = ( ) ( ) 75 15 75 15 2sin .cos 2 2 °+ ° °− ° = 2sin45°.cos30° = 1 3 2 2 2 × × = 3 2 112. If cosecθ + cotθ = 2, then the value of cosecθ is: (a) 3 2 (b) 2 (c) 5 4 (d) 5 SSC CGL 09/12/2022 (Shift-IV) Ans. (c): According to the question, cosecθ + cot θ = 2 ––––– (i) (cosec 2 θ – cot 2 θ) = 1 (cosecθ + cot θ) (cosecθ – cot θ) = 1 2(cosecθ – cot θ) = 1 (cosecθ – cot θ) = 1/2 …..(ii) ( ) 2 2 cosec cot 1 θ− θ= Q On solving Equation (i) and (ii), ⇒ 2cosecθ = 1 2 2 + ⇒ cosecθ = 5 4 113. If 5sin 2 A + 3cos 2 A = 4, 0<A<90°, then the value of tanA will be? (a) 0 (b) 2 (c) 3 (d) 1 SSC CGL 09/12/2022 (Shift-I) Ans. (d) : 5sin 2 A + 3cos 2 A = 4 5sin 2 A + 3(1 – sin 2 A) = 4 5sin 2 A + 3 – 3sin 2 A = 4 2sin 2 A = 1 sin 2 A = 1 2 ⇒ 1 sin A 2 = sinA = sin45° A = 45° ∴ tanA = tan45° = 1 114. The value of (cos 3 60° – cos 3 240º – cos 3 360º will be : (a) 9 7 − (b) 3 4 − (c) 7 5 − (d) 3 5 − SSC CGL 09/12/2022 (Shift-I) Ans. (b) : cos 3 60° – cos 3 240° – cos 3 360° = cos 3 60° – cos 3 (180°+60°) – cos 3 360° = cos 3 60° – (–cos 3 60°) – cos 3 360° = 3 3 1 1 1 2 2     − − −         = 1 1 1 8 8 + − = 6 8 − = 3 4 −

Trigonometry 124 . 115. If 2 sin θ 3sinθ + 2 = 0 − , then find the value of θ, where ≤ ≤ θ(0° θ 90°) . (a) 60° (b) 45° (c) 0° (d) 90° SSC CGL 12/12/2022 (Shift-II) Ans. (d) : 2 sin θ 3sinθ + 2 = 0 − ⇒ 2 sin θ 2sinθ sin 2=0 − − θ+ ⇒ ( ) ( ) sinθ sinθ 1 sin 2 =0 −2 − θ− ⇒ ( )( ) sin sin 2 0 θ−1 θ− = ⇒ sin , sin 2 θ=1 θ= 0 sin sin 90 θ= 0 90 θ= 116. ∆ABC is a right angle triangle whose right angle is at B and tanA = 3 4 , then the value of sinA + sin B + sin C will be : (a) 2 2 5 (b) 1 3 5 (c) 1 1 5 (d) 4 2 5 SSC CGL 12/12/2022 (Shift-II) SSC CGL 03/12/2022 (Shift-IV) Ans. (a) : Given that, 3 tan A 4 = = P B BC 3 tan A AB 4 = = By Pythagoras theorem AC 2 = BC 2 + AB 2 = 3 2 + 4 2 = 9 + 16 AC 2 = 25 AC 5 = 3 sin A , sin B sin 90 1 5 = = = , leLee 4 sin C 5 = ∴ 3 4 sin A sin B sin C 1 5 5 + + = + + ⇒ 2 2 5 117. Which of the following option is equal to 1 + sinθ cosθ , (where π θ 2 ≠ ) (a) 1 + cosθ sinθ (b) tanθ +1 tanθ 1 − (c) tanθ 1 tanθ 1 − + (d) cosθ 1 sin − θ SSC CGL 12/12/2022 (Shift-II) Ans. (d): 1 sin cos + θ θ where (θ ≠ π/2) = 1 sin 1 sin cos 1 sin + θ − θ × θ − θ ( ) 2 1 sin cos 1 sin − θ = θ − θ ( ) 2 cos cos 1 sin θ = θ − θ cos 1 sin θ = − θ 118. If secθ + θ + θ + θ + tanθ = θ = θ = θ = 5 , then find the value of (secθ θ θ θ – tanθ). θ). θ). θ). (a) 5 (b) 1 5 5 (c) 5 (d) 5 5 SSC CGL 12/12/2022 (Shift-III) Ans. (d) : Given that, secθ + tanθ = 5 Q sec 2 θ – tan 2 θ = 1 (secθ + tanθ) × (secθ – tanθ) = 1 5 × (secθ – tanθ) = 1 (secθ – tanθ) = 1 5 = 1 5 5 5 × 5 5 = 119. If 2 2sinθ + 2sin θ = 2, then the value of 2cos 4 θ + 2cos 2 θ) will be : (a) 0 (b) 1 (c) 4 (d) 2 SSC CGL 12/12/2022 (Shift-III) Ans. (d) : Given that, 2sinθ + 2sin 2 θ = 2 2(sinθ + sin 2 θ) = 2 sinθ + sin 2 θ = 1 sinθ = 1 – sin 2 θ sinθ = cos 2 θ sin 2 θ = cos 4 θ …. (i) Then 2cos 4 θ + 2cos 2 θ = 2sin 2 θ + 2cos 2 θ = 2(sin 2 θ + cos 2 θ) [จ sin 2 θ + cos 2 θ = 1] = 2 × 1 = 2 120. If θ is an acute angle and tanθ + cotθ = 2, then the value of (tan 200 θ + cot 200 θ) is : (a) 1 (b) –1 (c) 2 (d) 0 SSC CGL 13/12/2022 (Shift-I)

Trigonometry 125 . Ans. (c) : According to question- tanθ + cotθ = 2 Putting θ = 45º, tan45° + cot45° = 2 2 = 2 ∴ (tan45º) 200 + (cot45º) 200 = (1) 200 + (1) 200 = 1 + 1 = 2 121. If sec 2 A + tan 2 A = 3, then the value of cotA will be : (a) 3 (b) 0 (c) 1 3 (d) 1 SSC MTS 11/09/2023 (Shift II st ) Ans. (d) : According to the question, sec 2 A + tan 2 A = 3 1 + tan 2 A + tan 2 A = 3 2tan 2 A = 2 tan 2 A = 1 cot 2 A = 1 cotA = 1 122. If A is an acute angle, then the simplified form of the following will be? ( ) ( ) ( ) ( )             π cos π – A .cot + A cos –A 2 3π tan π + A tan + A sin 2π – A 2 is : (a) cos 2 A (b) cos 2 A (c) cos A (d) sin A SSC CGL 13/12/2022 (Shift-IV) Ans. (c) : ( ) ( ) ( ) ( ) cos A cot A cos A 2 3 tan A tan A sin 2 A 2 π   π− ⋅ + ⋅ −     π   π+ ⋅ + ⋅ π−     = ( ) ( ) ( ) ( ) ( ) 180 cos 180 A cot A cos A 2 tan 180 A tan 270 A sin 360 A   − ⋅ + ⋅ −     + ⋅ + ⋅ − ( ) ( ) ( ) cos A tan A cos A tan A cot A sin A − ×− × = − ×− = cos A cos A cos A sin A sin A × × = cosA 123. If sinθ + cosθ = 3-1 2 2 , then (tanθ + cotθ) is ? (a) 12(√3+2) (b) 8(√3+2) (c) 8(√3-2) (d) 12(√3-2) SSC CGL 05/12/2022 (Shift-I) Ans. (c) : sinθ + cosθ = 3 1 2 2 − On squaring both the sides, (sinθ + cosθ) 2 = 2 3 1 2 2   −           ⇒ sin 2 θ + cos 2 θ + 2sinθ.cosθ = 3 1 2 3 8 +− ⇒ 1 + 2sinθ. cosθ = 4 2 3 8 − ⇒ 2sinθ. cosθ = 4 2 3 8 8 − − ∴ sinθ. cosθ = ( ) 2 3 8 − + sin cos tan cot cos sin θ θ θ+ θ= + θ θ 2 2 sin cos 1 sin .cos sin .cos θ+ θ = θ θ θ θ 1 8 3–2 –(2 3) –(2 3) 3–2 8 = × + + 8( 3 – 2) = 124. 3 π π 16cos - 12cos = ________. 6 6 (a) 0 (b) 1 (c) –1 (d) 2 SSC CGL 01/12/2022 (Shift-IV) Ans. (a) : 3 16cos 12cos 6 6 π π − 3 0 0 3 16cos 30 12cos30 3 3 16 12 2 2 33 16 6 3 8 6 3 6 3 0 = −   = × − ×       = × − = − = 125. What is (sinα - sinβ) ? (a) 2cos sin 2 2 α−β α+β (b) 2sin sin 2 2 α+β α−β (c) 2cos sin 2 2 α+β α−β (d) 2cos cos 2 2 α+β α−β SSC CGL 01/12/2022 (Shift-II) Ans. (c) : sin C – sin D = 2cos C D 2 + . sin C D 2 − ∴ sin α – sin β = 2cos .sin 2 2 α+β α−β

Trigonometry 126 . 126. What will be the value of sin30º sin40º sin50º sin60º cos30º cos40º cos50ºcos60º . (a) 3 (b) 1 (c) 1 3 (d) 1 2 SSC CGL 01/12/2022 (Shift-II) Ans. (b) : = sin 30º.sin 40º.sin 50º.sin 60º cos30º.sin50º.sin 40º.cos60º [ ∵ cosθ = sin (90º– θ)] = 1 3 2 2 1 3 1 2 2 × = × 127. If cosec A – cotA = 1 4 , then the value of tan A is? (a) 8 17 (b) 8 15 (c) 15 17 (d) 17 15 SSC CGL 02/12/2022 (Shift-II) Ans. (b) : cosec A – cot A = 1 4 _____ (1) cosec 2 A – cot 2 A = 1 (cosec A + cot A) (cosec A – cot A) = 1 (cosec A + cot A) × 1 4 = 1 cosec A + cot A = 4 __________ (2) Equation (2) – Equation (1) cosec A + cot A = 4 cosec A – cot A 1 4 = – + – 15 2 cot A 4 = 8 tan A 15 = 128. If cotx – tanx = 3 2 , then the value of (cotx + tanx) will be : (a) 3 (b) 7/2 (c) 2 (d) 5/2 SSC CGL 02/12/2022 (Shift-I) Ans. (d) : ( ) ( ) 2 2 cot x tan x cot x tan x 4 cot x tan x + = − + 2 3 4 2   = +     1 cot tan   θ=   θ   Q 25 4 = 5 cot x tan x 2 + = 129. If sin(A + B) = cos(A + B), then the value of tanA will be : (a) 1 tan B 1 tan B + − (b) 1 sec B 1 sec B + − (c) 1 tan B 1 tan B − + (d) 1 cos ecB 1 cos ecB − + SSC CGL 02/12/2022 (Shift-I) Ans. (c) : sin (A + B) = cos (A + B) On dividing both sides by cos (A + B). ( ) tan A B 1 + = tan(A + B) = tan45° A + B = 45° A = 45° – B or, tanA = tan(45° – B) tan 45 tan B tan A 1 tan 45 tan B °− = + ° 1 tan B tan A 1 tan B − = + 130. If sec 2 A + tan 2 A = 4 7 , then the value of (sec 4 A – tan 4 A) will be : (a) 4 13 (b) 5 17 (c) 13 17 (d) 4 17 SSC CGL 02/12/2022 (Shift-I) Ans. (d) : sec 2 A + tan 2 A = 4 17 sec 4 A – tan 4 A = (sec 2 A + tan 2 A) (sec 2 A – tan 2 A) 2 2 4 1 sec A tan A 1 17   = × − =   Q = 4 17 131. − 1 + cosA 1 cosA is equals to : (a) sec A + tan A (b) cosec A + cot A (c) cosec A – cot A (d) sec A – tan A SSC CGL 03/12/2022 (Shift-IV) Ans. (b) : According to question, 2 2 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A (1 cos A) 1 cos A 1 cos A sin A 1 cos A sin A sin A cos ecA cot A + = − + + = × − + + = − + = = + = +

Trigonometry 127 . 132. If tan (A + B) = 3 and tan (A – B) = 1 3 , 0º < (A+B) < 90º then the value of A and B is respectively will be? (a) 45º and 15º (b) 15º and 45º (c) 30º and 30º (d) 60º and 30º SSC CGL (Mains) 02/03/2023 Ans. (a) : tan (A+B) = 3 tan (A+B) = tan 60º A+B = 60º –––––– (i) and, tan (A–B) = 1 3 tan (A–B) = tan30º A–B = 30º––––––– (ii) From equation (i) and (ii), A = 60º 30º 45º 2 + = B = 60° – 45° = 15° Hence A and B is 45° and 15° respectively. 133. If a cotθ + b cosecθ = p and b cotθ + a cosecθ = q then (p 2 – q 2 ) will be equal to? (a) b 2 - a 2 (b) b - a (c) a 2 + b 2 (d) a 2 - b 2 SSC CGL Mains (Tier-II) 07/03/2022 Ans. (a) : p 2 - q 2 = (acotθ+bcosecθ) 2 – (bcotθ+acosecθ) 2 = a 2 cot 2 θ+b 2 cosec 2 θ +2acotθ.bcosecθ–b 2 cot 2 θ– a 2 cosec 2 θ–2bcotθ.acosecθ = a 2 (cot 2 θ – cosec 2 θ) + b 2 (cosec 2 θ – cot 2 θ) = – a 2 + b 2 [∵ cosec 2 θ– cot 2 θ =1] = b 2 – a 2 134. (sin θ + cosec θ) 2 + (cos θ + sec θ) 2 = ? (a) 5 + tan 2 θ + cot 2 θ (b) 7 + tan 2 θ – cot 2 θ (c) 5 + tan 2 θ – cot 2 θ (d) 7 + tan 2 θ + cot 2 θ SSC CGL Mains (Tier-II) 07/03/2022 Ans. (d) : (sinθ + cosecθ) 2 + (cosθ + secθ) 2 2 2 2 2 2 2 sin cos sin .cos cos sec cos .sec ec ec = θ+ θ+ θ θ+ θ+ θ+ θ θ 2 2 2 2 2 2 (sin cos ) cos sec ec = θ+ θ+ θ+ θ+ + = 1 + 1 + cot 2 θ + 1 + tan 2 θ + 4 = 7 + tan 2 θ + cot 2 θ 135. Which of the following options is equal to       cosθ sinθ + sinθ cosθ ? (a) cotθ.secθ (b) secθ.tanθ (c) cosecθ.tanθ (d) cosecθ.secθ SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (d): cos sin sin cos θ θ + θ θ { } 2 2 2 2 cos sin cos sin 1 sin .cos θ+ θ = θ+ θ= θ θ Q 1 sin .cos = θ θ cos ec .sec = θ θ 136. Which of the following options is equal to       tanθ + secθ - 1 tanθ - secθ + 1 ? (a) 1 sin cos + θ θ (b) 1 tan cot + θ θ (c) 1 cos sin + θ θ (d) 1 cot tan + θ θ SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (a) : tan sec 1 tan sec 1 θ+ θ−     θ− θ+   ( ) { } 2 2 2 tan sec sec tan sec tan 1 tan sec 1 2   θ+ θ− θ− θ   = θ− θ= θ− θ+     Q 2 2 tan sec sec tan tan sec 1   θ+ θ− θ+ θ =   θ− θ+   2 2 tan sec – (sec – tan ) tan – sec 1 θ+ θ θ θ θ θ+ tan sec – (sec – tan )(sec tan ) tan – sec 1 θ+ θ θ θ θ+ θ θ θ+ ( )( ) tan sec 1 sec tan tan sec 1 θ+ θ − θ+ θ   =   θ− θ+   = (tanθ + secθ) sin 1 cos cos θ = + θ θ 1 sin cos + θ = θ 137. Which of the following is equal to (secA – cosA)? (a) tanA. sinA (b) cosA.sinA (c) cotA.cos (d) sinA.cotA SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (a) : secA – cosA 1 cos A cos A 1 = − 2 1 cos A cos A − = { } 2 2 2 sin A sin A 1 cos A cos A = = − Q sin A .sin A cos A = = tanA. sinA

Trigonometry 128 . 138. Which of the following is equal to       1 + tan tan θ θ ? (a) cosec sec θ θ (b) secθ. cosecθ (c) 1 (d) tan 2 θ SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (b) : 1 tan tan + θ θ 2 2 2 1 tan 1 tan sec tan + θ   = + θ= θ   θ Q 2 sec tan θ = θ 2 1/ cos sin / cos θ = θ θ 2 1 cos . cos sin θ = θ θ sec .cosec = θ θ 139. If 4sin 2 θ = 3(1 + cosθ) and 0 0 < θ < 90 0 , then find the value of (2 tanθ + 4sinθ – secθ). (a) 4 15 3 − (b) 3 15 4 − (c) 15 3 4 − (d) 15 3 3 + SSC CGL 11/04/2022 (Shift-I) Ans. (b) : 4sin 2 θ = 3 (1+cosθ), 0 0 < θ < 90 0 ( ) 2 3 sin 1 cos 4 θ= + θ ( ) 2 3 1 cos 1 cos 4 − θ= + θ ( )( ) ( ) 3 1 cos 1 cos 1 cos 4 + θ − θ= + θ 3 1 cos 4 − θ= 3 1 cos 1 4 4 θ= − = 1 cos 4 θ= 2 1 15 sin 1 cos 1 16 4 θ= − θ= − = sin 15 / 4 tan 15 cos 1/4 θ θ= = = θ ∴ 15 1 2 tan 4sin sec 2 15 4 4 1/4 θ+ θ− θ= × + × − 2 15 15 4 = + − 3 15 4 = − 140. Find the value of       2 1 - cotθ -1 1 - tanθ if 0 0 < θ < 90 0 . (a) cot 2 θ–1 (b) sec 2 θ+1 (c) sin 2 θ–1 (d) cos 2 θ–1 SSC CGL 12/04/2022 (Shift-II) Ans. (a) : 2 1 cot 1 1 tan − θ   −   − θ   2 cos 1 sin 1 sin 1 cos θ   −   θ = −   θ   − θ   2 sin cos sin 1 cos sin cos θ− θ     θ = −   θ− θ     θ   ( ) ( ) 2 2 2 2 sin cos cos 1 sin cos sin θ− θ θ = × − θ θ− θ ( ) ( ) 2 2 2 2 2 2 sin cos 2sin .cos cos 1 sin sin cos 2sin .cos θ+ θ− θ θ θ = × − θ θ+ θ− θ θ 2 2 cos 1 sin θ = − θ = cot 2 θ–1 141. If sec 2 θ + tan 2 θ = 1 3 2 , 0 0 < θ < 90 0 then the value of (cosθ + sinθ) will be : (a) 1 5 6 + (b) 1 5 3 + (c) 2 5 3 + (d) 9 2 5 6 + SSC CGL 18/04/2022 (Shift-I) Ans. (c) : 2 2 1 sec tan 3 2 θ+ θ= 2 2 7 sec sec 1 2 θ+ θ− = { } 2 2 tan sec 1 θ= θ− Q 2 7 2sec 1 2 θ= + 2 9 2sec 2 θ= 2 9 sec 4 θ= 3 sec 2 θ= 2 cos 3 θ= 2 4 5 5 sin 1 cos 1 9 9 3 θ= − θ= − = = 2 5 2 5 cos sin 3 3 3 + θ+ θ= + = 142. The value of 2 2 2 1 + cos – sin sec + cosec × sin (1 + cos tan + cot θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ) θ θ θ θ) θ θ θ θ) θ θ θ θ) θ θ , where→ 0° < θ < 90° is equal to : (a) cosec θ (b) cot θ (c) sec θ (d) tan θ SSC CGL (Tier-II) 29/01/2022 (Shift-I)

Trigonometry 129 . Ans. (b) : ( ) 2 2 2 1 cos sin sec cos ec sin 1 cos tan cot + θ− θ θ+ θ × θ + θ θ+ θ ( ) 2 2 2 1 1 cos 1 sin cos sin sin cos sin 1 cos cos sin + θ+ − θ θ θ = × θ θ θ + θ + θ θ ( ) 2 2 2 2 2 2 2 sin cos cos cos sin .cos sin 1 cos sin cos sin .cos θ+ θ θ+ θ θ θ = × θ + θ θ+ θ θ θ ( ) ( ) { } 2 2 1 cos 1 cos sin .cos sin cos 1 1 sin 1 cos sin .cos θ + θ θ θ = × θ+ θ= θ + θ θ θ cos sin θ = θ = cot θ 143. Which of the following is equal to       3 3 2 2 tan θ cot θ + + 2sinθ.cosθ sec θ cosec θ ÷(1+ cosec 2 θ + tan 2 θ), where, 0 0 < θ < 90 0 (a) cosecθ secθ (b) sinθ cosθ (c) sec θ (d) cosec θ SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (b) : ( ) 3 3 2 2 2 2 tan θ cot θ + + 2sinθcosθ ÷ 1+ cosec θ + tan θ sec θ cosec θ       ( ) 3 2 3 2 2 2 3 3 sin cos cos sin 2sin .cos sec cos ec 1 1 cos sin   θ θ θ θ = × + × + θ θ ÷ θ+ θ   θ θ   3 3 2 2 sin cos 1 1 = 2sin cos cos sin cos sin   θ θ   + + θ θ÷ +     θ θ θ θ     4 4 2 2 2 2 sin cos sin cos 2sin cos cos .sin cos .sin   θ+ θ θ+ θ = + θ θ ÷   θ θ θ θ   ( ) 2 2 2 2 2 sin cos cos sin cos sin 1 θ+ θ θ θ = × θ θ 2 1 cos sin cos sin 1 2 θ θ = × θ θ sin .cos = θ θ 144. Suppose that 0 0 < θ < 90 0 , which of the following is equal to (1 + cot 2 θ) (1+tan 2 θ) × (sinθ – cosec θ) (cosθ – secθ) (a) sinθ cosθ (b) secθ + cosecθ (c) sin θ + cos θ (d) sec θ cosec θ SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (d) : (1 + cot 2 θ) (1+tan 2 θ) × (sinθ–cosecθ) (cosθ–secθ) sin 1 1 tan ,cosec ,sec cos sin cos θ   ∴ θ= θ= θ=   θ θ θ   2 2 2 2 cos sin 1 1 1 1 sin . cos sin cos sin cos    θ θ    = + + × θ− θ−       θ θ θ θ       ( ) ( ) 2 2 2 2 sin 1 cos 1 1 1 . sin cos sin cos θ− θ− = × × θ θ θ θ (∴ sin 2 θ + cos 2 θ = 1 Ùee cos 2 θ =1 – sin 2 θ) 2 2 2 2 1 1 cos sin sin cos sin cos θ θ = × × × θ θ θ θ 1 sin cos = θ θ = secθ.cosecθ 145. If A = 30°, then the following is equal to? 2 [6sinA + 9cosecA - cot A] 12sinA (a) 3 (b) –3 (c) –6 (d) 6 SSC CHSL –06/06/2022 (Shift-III) Ans. (a) : Given that, A = 30º then 2 6sinA + 9cosecA - cot A 12sinA 2 6sin30º +9cosec30º-cot 30º 12sin30º = 1 6 9 2 3 2 1 12 2 × + × − = × 3 18 3 3 6 + − = = 146. (cosec θ – cot θ θθ θ) 2 = ?, 0º < θ < 90º ––––––– (a) 1 cos 1 cos − θ + θ (b) 1 cos 1 cos + θ − θ (c) 1 sin 1 cos − θ + θ (d) 1 cos 1 sin + θ − θ SSC CHSL –09/06/2022 (Shift-III) Ans. (a) : (cosec θ – cot θ) 2 = cosec 2 θ + cot 2 θ –2 cosec θ . cot θ 2 2 2 1 cos 1 cos 2 . sin sin sin sin θ θ = + − θ θ θ θ 2 2 2 1 cos 2cos sin sin + θ θ = − θ θ 2 2 1 cos 2cos sin + θ− θ = θ 2 2 (1 cos ) 1 cos − θ = − θ (1 cos )(1 cos ) (1 cos )(1 cos ) − θ − θ = − θ + θ 1 cos 1 cos − θ = + θ 147. If sinθ = 12 13 , 0° < θ < 90°, then find the value of 2 2 2 sin θ-cos θ 1 × 2sinθ.cosθ tan θ . (a) 695 3542 (b) 595 3456 (c) 295 3456 (d) 590 3542 SSC CHSL –08/06/2022 (Shift-II)

Trigonometry 130 . Ans. (b): Given that 12 AB sin 13 AC θ= = 2 2 BC 13 12 = − 169 144 25 5 = − = = 5 cos 13 θ= Hence, 2 2 2 2 2 2 2 sin cos 1 sin cos cos 2sin .cos tan 2sin cos sin θ− θ θ− θ θ × = × θ θ θ θ θ θ 2 2 2 2 12 5 5 13 13 13 12 5 12 2 13 13 13       −             = ×       ×             144 25 25 169 169 169 24 5 144 169 169 − = × × 119 25 595 24 5 144 3456 = × = × 148. What will be the value of sin4θ (1 - cos4θ) ? (a) cot θ (b) cot 2θ (c) tan 2θ (d) tan θ SSC CHSL –01/06/2022 (Shift-III) Ans. (b) : ∵ sin 2θ = 2 sin θ . cos θ cos 2 θ = 2 cos 2 θ – 1 = 1 – 2 sin 2 θ ∴ According to the question, sin4θ (1- cos4θ) ( ) 2 2sin2 .cos2 1 1 2sin 2 θ θ = − − θ 2 2sin2 .cos2 2sin 2 θ θ = θ cos 2 cot 2 sin 2 θ = = θ θ 149. Simplify the following expression: cosA sinA + – sinA 1 – tanA 1 – cotA (a) 1+cosA (b) (1 + sin A ) CosA (c) 1 + sinA (d) cos A SSC CGL (Tier-I) 19/04/2022 (Shift-III) Ans. (d) cos A sin A sin A 1 tan A 1 cot A + − − − cos A sin A sin A sin A cos A 1 1 cos A sin A = + − − − 2 2 cos A sin A sin A cos A sin A sin A cos A = + − − − ( ) 2 2 cos A sin A sin A cos A sin A − = − − = cosA + sinA – sinA = cosA 150. The value of cotθ + cosθ 2- , cotθ - cosθ when 0º < θ < 90º is equal to: (a) 2 + sec θ + tan θ (b) 2 – secθ + tan θ (c) 2 – sec θ – tan θ (d) 2 + secθ – tan θ SSC CGL (Tier-I) 18/04/2022 (Shift-II) Ans. (b) ( ) 2 2 2 cot cos cotθ + cosθ 2– 2 cotθ - cosθ cot cos θ+ θ = − θ− θ ( ) 2 2 2 cos 1 sin (cot cos ) sin 2 2 cos cos cos cos sin sin θ + θ θ+ θ θ = − = − θ   θ × θ − θ   θ θ   =2–secθ+tanθ 151. If , 1 2 2 sec θ + tan θ = 3 , 0º < θ < 90º 2 then (cos θ + sin θ ) is equal to. (a) 1 5 3 + (b) 2 5 3 + (c) 1 5 6 + (d) 9 2 5 6 + SSC CGL (Tier-I) 18/04/2022 (Shift-I) Ans. (b) 2 θ 1 7 2 sec θ + tan =3 = 2 2 ( ) 2 2 sec tan 1 θ− θ= Q 2 2 2 sec tan 1 9 2sec 2 θ− = θ= = 2 sec 3 2 or,cos 3 θ= θ= 2 4 5 sin 1 cos 1 9 3 ∴ θ − θ= − = 2 5 2 5 cos sin 3 2 3 + ∴ θ+ θ= = = 152. Simplify          2 1 1 sec α 1+ 1- cosecα cosecα . (a) tan 4 α (b) sin 2 α (c) 1 (d) –1 SSC CGL–(Tier-I) 24/08/2021 (Shift I) Ans. (c) : 2 1 1 sec 1 1 cosec cosec    α + −    α α    = sec 2 α. (1+sinα) (1–sinα) = sec 2 α. (1–sin 2 α) ( ) 2 2 1 Sin cos − α= α Q = sec 2 α. cos 2 α 2 2 1 .cos 1 cos = α= α 153. For < θ < 90 0 , 1 1 + cosθ tanθ - secθ is equal to : (a) sec θ (b) –sec θ (c) –tan θ (d) tan θ SSC CGL–(Tier-I) 17/08/2021 (Shift II)

Trigonometry 131 . Ans. (c) : 1 1 + cosθ tanθ - secθ ( )( ) sec tan sec sec tan sec tan θ+ θ = + − θ− θ θ+ θ 2 2 2 2 sec tan sec sec tan 1 sec tan θ+ θ = θ− θ− =     θ− θ Q = secθ– secθ – tanθ = – tanθ 154. - - (1 + cos θ) (cosec θ cot θ) sec θ =? sin θ (1 sin θ) (sec θ + tan θ) (a) sec 2 θ (b) sin 2 θ (c) cos 2 θ (d) cosec 2 θ SSC CHSL 19/04/2021 (Shift-I) Ans. (a) : (1+ cos θ) (cosec θ cot θ) sec θ =? sin θ (1 sin θ) (sec θ + tan θ) - - = (1 cos ) (1 cos ) sec sin (1 sin ) sin (1 sin ) cos − θ + θ × θ θ + θ θ − θ θ = 2 2 1 sin cos sin 1 cos sin cos θ θ θ θ× θ θ = 2 2 2 sin 1 cos sin θ × θ θ = sec 2 θ 155. If cosec 2 θ (cos θ θθ θ – 1)(1 + cos θ θθ θ) = k, then what is the value of k? (a) 1 (b) –1 (c) ½ (d) 0 SSC CHSL 06/08/2021 (Shift-I) Ans. (b) : cosec 2 θ (cos θ – 1)(1 + cos θ) = k 2 1 sin − θ × (1–cosθ) (1+cosθ) = k 2 2 sin sin − θ θ = k k = –1 156. If - cosecθ + cotθ cosecθ cotθ = 7, then the value of - 2 2 4 sin θ+5 4 sin θ 1 is: (a) 3 (b) 15 (c) 9 (d) 12 SSC CHSL 13/04/2021 (Shift-I) Ans. (c) : cosec cot 7 cosec cot 1 θ+ θ = θ− θ By componendo and Dividendo method 2 cosec 7 1 2 cot 7 1 θ + = θ − 1 8 sin cos 6 sin θ = θ θ 1 4 cos 3 = θ cos θ 3 4 = Q sin 2 θ = 1–cos 2 θ = 9 7 1 16 16 − = 2 2 7 5 4sin 5 27 4 9 7 3 4sin 1 1 4 + θ+ = = = θ− − 157. If , sec θ + tan θ = 3 then the value of sec θ is: (a) 4 3 (b) 3 4 (c) 5 3 (d) 3 5 SSC CGL (Tier-II)– 18/11/2020 Ans. (c) : Secθ + tanθ = 3 ________ (1) Q Sec 2 θ – tan 2 θ = 1 (Secθ + tanθ) (Secθ – tanθ) = 1 3(Secθ – tanθ) = 1 Secθ – tanθ 1 3 = _______ (2) From equation (i) and (ii) 5 Sec 3 θ= 158. If secθ + tanθ 51 =2 secθ - tanθ 79 then the value of sin θ is equal to : (a) 91 144 (b) 39 72 (c) 65 144 (d) 35 72 SSC CGL (Tier-II) – 18/11/2020 Ans. (c) : Sec tan 51 2 Sec tan 79 θ+ θ = θ− θ By componendo and dividendo method Sec tan Sec tan 209 79 Sec tan Sec tan 209 79 θ+ θ+ θ− θ + = θ+ θ− θ+ θ − Sec 288 tan 130 θ = θ 1 144 Sin 65 = θ Sin θ 65 144 =

Trigonometry 132 . 159. If secθ – tanθ = P, then cosecθ = ? (a) 2 2P 1 P − (b) 2 2 P 1 1 P + − (c) 2 2 1 P 1 P − + (d) 2 2P 1 P + SSC CGL (TIER-I)– 04.06.2019 (Shift-III) Ans. (b) : secθ – tanθ = P then, 2 2 1 sec tan sec tan 1 P   θ+ θ= θ− θ=   Q 2secθ 1 P P = + 2secθ 2 P 1 P + = secθ 2 P 1 2P + = 2 2 P 1 cosec 1 P + θ= − 160. 2 3 o 3 sin 2sin 1, 45 , 2cos cos   θ− θ + θ≠     θ− θ   is equal to: (a) sec 2 θ (b) 2tan 2 θ (c) cosec 2 θ (d) cot 2 θ SSC CGL (TIER-I)– 04.06.2019 (Shift-II) Ans. (a) : 2 3 o 3 sin 2sin 1, 45 2cos cos   θ− θ + θ≠     θ− θ   2 2 2 sin 1 2sin 1 cos 2cos 1     θ − θ = +       θ θ−       2 cos 2 tan 1 cos 2   θ   = θ +     θ     [ ] 2 tan 1 = θ + = tan 2 θ + 1 = sec 2 θ 161. . 2 2 2 + tan θ + cot θ secθ cosecθ is equal to: (a) cosθ sinθ (b) secθ cosecθ (c) tan θ (d) cot θ SSC CGL (TIER-I) – 04.06.2019 (Shift-I) Ans. (b) : 2 2 2 tan cot ? sec .cosec + θ+ θ = θ θ 2 2 1 tan 1 cot sec .cosec + θ+ + θ = θ θ 2 2 sec cosec sec .cosec θ+ θ = θ θ sec cosec cosec sec θ θ = + θ θ sin cos cos sin θ θ = + θ θ 2 2 sin cos sin .cos θ+ θ = θ θ 1 sin .cos = θ θ sec .cosec = θ θ 162. The value of 2 1 cot sin 1 cosec θ − θ + θ θ + θ θ + θ θ + θ is: (a) 1 (b) –1 (c) 0 (d) 2 SSC CGL (TIER-I)– 06.06.2019 (Shift-III) Ans. (a) : 2 1 cot sin 1 cosec θ − θ + θ 2 1 cosec 1 sin cosec 1 θ− = − θ θ+ ( ) cosec cosec 1 = θ− θ− = 1 163. If tan cot 1 k, 1 cot 1 tan θ θ θ θ θ θ θ θ + = + + = + + = + + = + − θ − θ − θ − θ − θ − θ − θ − θ then k = ____. (a) tanθ + secθ (b) tanθ cosecθ (c) cotθ + secθ (d) cosecθ secθ SSC CGL (TIER-I) – 07.06.2019 (Shift-III) Ans. (d) : tan cot 1 k 1 cot 1 tan θ θ + = + − θ − θ sin / cos cos / sin 1 k 1 cos / sin 1 sin / cos θ θ θ θ + = + − θ θ − θ θ ( ) ( ) 2 2 sin cos 1 k cos sin cos sin cos sin θ θ ⇒ + = + θ θ− θ θ θ− θ ( ) 3 3 sin cos 1 k sin .cos . sin cos θ− θ ⇒ = + θ θ θ− θ ( ) ( ) ( ) 2 2 sin cos sin cos sin .cos 1 k sin .cos . sin cos θ− θ θ+ θ+ θ θ ⇒ = + θ θ θ− θ ( ) 1 sin .cos 1 k sin .cos + θ θ ⇒ = + θ θ

Trigonometry 133 . 1 sin .cos 1 k sin .cos sin .cos θ θ ⇒ + = + θ θ θ θ cosec .sec 1 1 k ⇒ θ θ+ = + By compare the both side k = cosec .sec θ θ 164. If tan sec 1 1 sec , tan sec 1 k θ− θ+ θ− θ+ θ− θ+ θ− θ+         θ= θ= θ= θ=         θ+ θ− θ+ θ− θ+ θ− θ+ θ−         then k = ____. (a) 1– cosθ (b) 1– sinθ (c) 1+ cosθ (d) 1 + sin θ SSC CGL (TIER-I) – 07.06.2019 (Shift-II) Ans. (d) : tan sec 1 1 .sec tan sec 1 k θ− θ+   θ=   θ+ θ−   2 2 sec θ tan θ 1   − =   Q 2 2 tan sec sec tan 1 .sec tan sec 1 k   θ− θ+ θ− θ θ=   θ+ θ−   ( ) [ ] ( ) sec tan 1 sec tan 1 .sec tan sec 1 k θ− θ −+ θ+ θ θ= θ+ θ− 1 sin 1 1 . cos cos cos k θ   − =   θ θ θ   2 1 sin 1 k cos − θ = θ 2 1 sin 1 k 1 sin − θ = − θ 1 1 1 sin k = + θ k 1 sin = + θ 165. The value of 1 1 , sec x tan x cos x − − 0 0 < x < 90 0 , is equal to: (a) tan x (b) 2sec x (c) cot x (d) 2cos x SSC CGL (TIER-I) – 07.06.2019 (Shift-I) Ans. (a) : 1 1 sec x tan x cos x − − ( )( ) sec x tan x 1 sec x tan x sec x tan x cos x + ⇒ − − + 2 2 sec x tan x 1 cos x sec x tan x + ⇒ − − We know that, 2 2 sec x tan x 1 − = sec x tan x sec x 1 + ⇒ − sec x tan x sec x ⇒ + − tan x ⇒ 166. If tan sin k 1 tan sin k 1 θ+ θ + θ+ θ + θ+ θ + θ+ θ + = θ− θ − θ− θ − θ− θ − θ− θ − , then k = ? (a) sin θ (b) cos θ (c) sec θ (d) cosec θ SSC CGL (TIER-I) – 10.06.2019 (Shift-I) Ans. (c) : tan sin k 1 tan sin k 1 θ+ θ + = θ− θ − By componendo and Dividendo method ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) tan sin tan sin k 1 k 1 tan sin tan sin k 1 k 1 θ+ θ+ θ− θ + + − = θ+ θ− θ− θ + − − tan k 1 k sec sin 1 cos θ = ⇒ = = θ θ θ 167. What is the value of [(cos 7A+cos 5A) ÷ sin 7A– sin 5A)]? (a) tan A (b) tan 4 A (c) cot 4 A (d) cot A SSC CGL (Tier-II) 19-02-2018 Ans. (d) : cos7A cos5A sin 7A sin 5A + − C D C D cos C cos D 2cos .cos 2 2 + − + = ∵ sin C - sin D = C D C D 2cos .sin 2 2 + − 12A 2A 2cos .cos 2 2 12A 2A 2cos .sin 2 2 ∴ = cot A 168. What is the value of ( ) ( )     2 1 sec2θ + 1 sec θ-1 × cotθ - tanθ 2 ? (a) 0 (b) 1 (c) cosecθ (d) sec θ SSC CGL (Tier-II) 18-02-2018 Ans. (b) : ( ) ( ) ( ) 2 1 sec 2 1 sec 1 cot tan 2   θ+ θ− × θ− θ     ( ) 1 cos sin sec 2 1 tan 2 sin cos  θ θ   θ+ θ× −     θ θ     2 2 1 1 cos sin 1 tan cos 2 2 sin .cos     θ− θ   + θ×       θ θ θ       2 1 cos 2 1 tan 2 cos 1 sin 2 θ   + θ×   θ− θ   sin 2 cos 2 cos 2 sin 2 θ θ × θ θ = 1 169. The value of ( ) 2 2 tan θ cosec θ-1 sinθ + cosθ - 1 × sinθ - cosθ + 1 secθ - tanθ is :

Trigonometry 134 . (a) 1 (b) 0 (c) –1 (d) 1 2 SSC CGL (Tier-II) 13-09-2019 Ans. (a) : ( ) ( ) ( ) 2 2 tan cosec 1 sin cos 1 sin cos 1 sec tan θ θ− θ+ θ− = × θ− θ+ θ− θ ( ) ( ) ( ) 2 2 sin cos 1 tan .cot .cos 1 sin cos 1 sin θ+ θ− θ θ θ = × + θ− θ − θ ( ) 2 2 sin .cos cos cos 1 sin cos cos .sin θ θ+ θ− θ = − θ− θ+ θ θ 2 2 sin .cos cos cos cos cos cos .sin θ θ+ θ− θ = θ− θ+ θ θ = 1 170. The value of ( )( ) sinθ - cosθ 1 + tanθ + cotθ =? 1 + sinθcosθ (a) secθ–cosecθ (b) cosecθ–secθ (c) tanθ–cotθ (d) sinθ+cosθ SSC CGL (Tier-II) 13-09-2019 Ans. (a) : ( )( ) sin cos 1 tan cot 1 sin .cos θ− θ + θ+ θ + θ θ ( ) sin cos sin cos 1 cos sin 1 sin .cos θ θ   θ− θ + +   θ θ   = + θ θ ( ) ( ) 2 2 sin .cos sin cos sin cos cos .sin 1 sin .cos θ θ+ θ+ θ θ− θ× θ θ = + θ θ ( )( ) ( ) sin cos 1 sin .cos 1 sin .cos cos .sin θ− θ + θ θ = + θ θ θ θ sin cos cos .sin cos .sin θ θ = − θ θ θ θ = secθ – cosecθ 171. If secθ + tanθ = p, (p > 1) then cosecθ + 1 =? cosecθ - 1 (a) 2p 2 (b) p 1 p 1 + − (c) p 2 (d) p 1 p 1 − + SSC CGL (Tier-II) 13-09-2019 Ans. (c) : secθ+tanθ = p 1 sin p cos cos θ + = θ θ , 1 sin p cos + θ = θ ( ) 2 2 2 1 sin p cos + θ = θ ( ) ( )( ) 2 2 1 sin p 1 sin 1 sin + θ = − θ + θ 2 1 sin p 1 sin + θ = − θ 2 cosec 1 p cosec 1 θ+ = θ− 172. The value of (1+ cotθ – cosecθ) (1+cosθ+sinθ) secθ = ? (a) sinθcosθ (b) 2 (c) –2 (d) secθcosecθ SSC CGL (Tier-II) 13-09-2019 Ans. (b) : (1+cotθ–cosecθ) (1+cosθ+sinθ). secθ ( ) cos 1 1 1 1 cos sin sin sin cos θ   = + − × + θ+ θ×   θ θ θ   ( )( ) sin cos 1 sin cos 1 sin .cos θ+ θ− θ+ θ+ = θ θ ( ) 2 sin cos 1 sin .cos θ+ θ − = θ θ ( ) 2 2 sin cos 2sin .cos 1 sin .cos θ+ θ+ θ θ− = θ θ 2 2 2sin .cos sin cos 1 sin .cos θ θ   = θ+ θ=   θ θ Q = 2 173. The value of ( )( )( ) ( ) ( ) sec 1-sin sin + cos sec + tan sin 1 + tan + cos 1 + cot φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ is equal to : (a) 2cosφ (b) 2sinφ (c) cosecφsecφ (d) sinφcosφ SSC CGL (Tier-II) 12-09-2019 Ans. (d) : ( )( )( ) ( ) ( ) sec 1 sin sin cos sec tan sin 1 tan cos 1 cot φ − φ φ+ φ φ+ φ φ + φ+ φ + φ ( )( )( ) sec tan sin cos sec tan cos sin sin cos sin cos cos sin φ− φ φ+ φ φ+ φ     φ+ φ φ+ φ φ + φ     φ φ     ( ) ( ) ( )( ) 2 2 sec tan sin cos sin cos tan cot φ− φ φ+ φ = φ+ φ φ+ φ ( ) 2 2 1 sin cos sin cos sin cos sin cos cos sin φ φ = = = φ φ   φ φ φ+ φ +   φ φ   174. The value of cosec - cot sin ÷ cosec + cot 1 + cos φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ is equal to : (a) cosec φ (b) 1 2 (c) 1 (d) sec φ SSC CGL (Tier-II) 12-09-2019

Trigonometry 135 . Ans. (c) : cosec cot sin cosec cot 1 cos φ− φ φ = ÷ φ+ φ + φ 1 cos sin 1 cos 1 cos   − φ φ = ÷   + φ + φ   ( ) 2 2 1 cos sin sin 1 cos − φ   φ = ÷   φ + φ   ( ) ( ) 1 cos 1 cos sin sin − φ + φ = × φ φ ( ) 2 2 1 cos sin − φ = φ 2 2 sin 1 sin φ = = φ 175. ( ( ( ) ) )( ( ( ) ) ) 2sinA 1 sin A 1 sin A cos A + + + + + + + + + is equal to : (a) 1 + sin A cos A (b) 1 + sin A – cos A (c) 1 + cos A – sin A (d) 1 – sin A cos A SSC CGL (Tier-II) 11-9-2019 Ans. (b) : ( )( ) 2sinA 1 sin A 1 sin A cos A + + + ( ) 2sinA 1 sin A (1 sin A) cos A (1 sin A) cos A (1 sin A) cos A + + − = × + + + − ( ) [ ] 2 2 2sinA 1 sin A (1 sin A) cos A 1 sin A 2sinA cos A + + − = + + − ( ) [ ] 2 2sinA 1 sin A 1 sin A cos A 2sin A 2sinA + + − = + ( ) ( ) ( ) 2 2 2sinA 2sin A 1 sin A cos A 2sin A 2sinA + + − = + 1 sin A cos A = + − 176. – cot cos cot cos θ+ θ θ+ θ θ+ θ θ+ θ θ θ θ θ θ θ θ θ is equal to : (a) 1–sec θ tan θ (b) sec θ + tan θ (c) sec θ – tan θ (d) 1 + sec θ tan θ SSC CGL (Tier-II) 11-9-2019 Ans. (b) : cot cos cot cos θ+ θ θ− θ = 1 1 tan sec 1 1 tan sec + θ θ − θ θ = (sec tan ) (sec tan ) (sec tan ) (sec tan ) θ+ θ θ+ θ × θ− θ θ+ θ = ( ) ( ) 2 2 2 2 2 sec + tan sec tan 1 sec tan θ θ ∴ θ− θ= θ− θ = secθ + tanθ 177. If cosx = p q and 0 0 < x < 90 0 , then the value of tanx is: (a) 2 2 q p q − (b) 2 2 q q p − (c) 2 2 p p q − (d) 2 2 q p p − SSC CHSL –15/10/2020 (Shift-I) Ans. (d) : cosx = P q sec x q p → = 1 cos sec   θ=   θ   Q sec 2 x = 2 2 q p ∴ tanx = 2 2 2 q sec x 1 1 P − = − = 2 2 q P P − 178. If sinθ – cosθ = 1 29 find the value of sinθ + cosθ. (a) 2 29 (b) 22 29 (c) 42 29 (d) 41 29 SSC CHSL –18/03/2020 (Shift-II) Ans. (d) : sinθ – cosθ 1 29 = ⇒ Height Base 1 – = Hypotenuse Hypotenuse 29 21 20 29 29 = − From triplet, 20, 21, 29 ∴ sinθ + cosθ = Height Base + Hypotenuse Hypotenuse 21 20 41 = 29 29 29 + = 179. If 0 0 < θ < 90 0 then the value of (( ( ) )( ( ( ) ) ( ) 2 sec 1 sin sec tan sec tan θ − θ θ+ θ θ − θ θ+ θ θ − θ θ+ θ θ − θ θ+ θ θ− θ θ− θ θ− θ θ− θ 1 k 1 k + = − , k is equal : (a) cosθ (b) sinθ (c) cosecθ (d) secθ SSC CHSL 04/07/2019 (Shift-I) Ans. (b) : 0 0 < θ < 90 0 ( )( ) ( ) 2 sec 1 sin sec tan 1 k 1 k sec tan θ − θ θ+ θ + ⇒ = − θ− θ ( ) ( ) 2 2 1 1 sin 1 sin 1 k cos cos 1 k 1 sin cos + θ   − θ   + θ θ   ⇒ = − − θ θ ( )( ) ( ) 2 1 sin 1 sin 1 k 1 k 1 sin − θ + θ + ⇒ = − − θ

Trigonometry 136 . 1 sin 1 k 1 sin 1 k + θ + ⇒ = − θ − then, k sin = θ 180. If 1 1 sec k, sec tan cos − = θ× − = θ× − = θ× − = θ× θ− θ θ θ− θ θ θ− θ θ θ− θ θ 0 0 < θ < 90 0 then K is equal : (a) cosecθ (b) cotθ (c) tanθ (d) sinθ SSC CHSL 02/07/2019 (Shift-II) Ans. (d) : 0 0 1 1 sec k, 0 90 sec tan cos − = θ× <θ< θ− θ θ 1 1 sec k 1 sin cos cos cos − = θ× θ θ − θ θ cos 1 sec k 1 sin cos θ − = θ× − θ θ ( ) ( ) 2 cos 1 sin sec k cos 1 sin θ− − θ = θ× θ − θ ( ) ( ) 2 2 2 cos 1 sin sec k cos 1 sin cos 1 sin θ− + θ = θ× θ− =− θ θ − θ Q ( ) 2 sin sin sec k cos 1 sin − θ+ θ = θ× θ − θ ( ) ( ) sin sin 1 sec k cos 1 sin θ− θ+ = θ× θ − θ sin k cos cos θ = θ θ k sin = θ 181. Simplify sin 1 cos 1 1 cos sin tan cot θ + θ θ + θ θ + θ θ + θ             +             + θ θ θ+ θ + θ θ θ+ θ + θ θ θ+ θ + θ θ θ+ θ             (a) 2sinθ (b) 2cosθ (c) cosθ (d) sinθ SSC CHSL 02/07/2019 (Shift-III) Ans. (b) : sin 1 cos 1 1 cos sin tan cot θ + θ    +    + θ θ θ+ θ    ( ) 2 2 sin 1 cos 2 cos 1 sin cos 1 cos .sin cos sin       θ+ + θ+ θ =      θ θ + θ θ     +   θ θ   ( ) 2 2 2 2 sin cos 1 2cos cos .sin 1 cos sin sin cos   θ+ θ+ + θ θ θ   =       + θ θ θ+ θ     ( ) 2 2 1 1 2 cos cos .sin 1 cos .sin sin cos   + + θ θ θ   =       + θ θ θ+ θ     ( ) ( ) ( ) 21 cos cos .sin 1 cos sin 1 + θ θ θ = + θ θ 2 cos = θ 182. ( )( ) cotθ 1 - sinθ secθ + tanθ is equal : (a) cosec θ (b) sin θ (c) sec θ (d) 1 SSC CHSL 01/07/2019 (Shift-III) Ans. (a) ( )( ) ( ) cos cot sin 1 sin 1 sin sec tan 1 sin cos cos θ θ θ = θ − θ θ+ θ   − θ +   θ θ   ( ) cos sin 1 sin 1 sin cos θ θ = + θ   − θ   θ   2 cos sin 1 sin cos θ θ = − θ θ 2 cos sin cos cos θ θ = θ θ 1 sin = θ = cosecθ 183. ( ( ( )) )( ( ( ) ) ) ( )( ) sec tan 1 sin cosec 1 cos cosec cot θ+ θ − θ θ+ θ − θ θ+ θ − θ θ+ θ − θ θ + θ θ− θ θ + θ θ− θ θ + θ θ− θ θ + θ θ− θ equal is : (a) secθ (b) sinθ (c) cosθ (d) cosecθ SSC CHSL 02/07/2019 (Shift-I) Ans. (c) : ( )( ) ( )( ) sec tan 1 sin cosec 1 cos cosec cot θ+ θ − θ θ + θ θ− θ ( ) ( )( ) 1 sin 1 sin cos cos 1 1 cos cosec cot sin θ   + − θ   θ θ   = + θ θ− θ θ ( ) ( )( ) 1 sin 1 sin cos cosec cot cosec cot + θ   − θ   θ   = θ+ θ θ− θ 2 2 2 1 sin cos cosec cot − θ θ = θ− θ 2 cos cos cos θ = = θ θ 2 2 cosec cot 1   θ− θ=   Q 184. If 1 cos θ cosec θ cot θ 1 r 1 cosθ cosecθ cot θ 1 r − − − − − − − − − − − − × = × = × = × = + + + + + + + + + + + + then the value of r is : (a) sinθ (b) cosecθ (c) cosθ (d) secθ SSC CHSL 03/07/2019 (Shift-II)

Trigonometry 137 . Ans. (c) : 1 cos θ cosecθ cot θ 1 r 1 cos θ cosecθ cot θ 1 r − − − × = + + + 1 cos θ 1 cosθ 1 r sin θ sin θ 1 cos θ 1 cos θ 1 r sin θ sin θ − − − ⇒ × = + + + 1 cosθ 1 cosθ 1 r sin θ 1 cos θ 1 cos θ 1 r sin θ − − − ⇒ × = + + + 1 cosθ 1 cos θ 1 r 1 cos θ 1 cos θ 1 r − − − ⇒ × = + + + 1 cos θ 1 r 1 cosθ 1 r − − ⇒ = + + Hence, it is clear that r = cosθ 185. If ( ) ( ) 3 2 2 sin cos ec cos sec r tan sin θ− θ θ− θ θ− θ θ− θ θ− θ θ− θ θ− θ θ− θ = θ− θ θ− θ θ− θ θ− θ then the value of r is : (a) tanθ (b) cosecθ secθ (c) cotθ (d) sinθ cosθ SSC CHSL 03/07/2019 (Shift-I) Ans. (c) : ( )( ) 3 2 2 sin cosec cos sec r tan sin θ− θ θ− θ = θ− θ 2 2 2 1 1 sin cos sin cos sin sin cos    θ− θ−    θ θ    = θ − θ θ 2 2 2 2 sin 1 cos 1 sin cos 1 sin 1 cos    θ− θ−    θ θ    =   θ −   θ   ( ) ( ) 2 2 2 2 2 cos sin sin .cos 1 cos sin cos − θ×− θ θ θ =   − θ θ   θ   2 2 2 2 2 cos .sin .cos sin .cos .sin .sin θ θ θ = θ θ θ θ 3 3 3 cos cos sin sin θ θ   = =   θ θ   , r cot = θ 186. If 2 2p cos p 1 θ= θ= θ= θ= + then sinθ is : (a) 2 2p p 1 + (b) 2 2 p 1 p 1 + − (c) 2 2 p 1 p 1 − + (d) 2 2p p 1 − SSC CHSL 04/07/2019 (Shift-II) Ans. (c) : 2 2p cos p 1 θ= + ( ) ( ) ( ) 2 2 2 AB AC BC = − ( ) ( ) 2 2 2 p 1 2p = + − 4 2 2 p 1 2p 4p = + + − 4 2 p 1 2p = + − ( ) ( ) 2 2 2 AB p 1 = − 2 AB p 1 = − 2 2 AB p 1 sin AC p 1 − θ= = + 187. If 12cot 2 θ – 31 cosecθ + 32 = 0, 0º < θ < 90º then the value of sin θ is : (a) 43 , 54 (b) 54 , 43 (c) 3 , 32 1 (d) 21 , 34 SSC CHSL 05/07/2019 (Shift-I) Ans. (a) : 2 12cot θ 31cosecθ 32 0 − + = 2 12(cosec θ 1) 31cosecθ 32 0 − − + = 2 12cosec θ 12 31cosecθ 32 0 − − + = 2 12cosec θ 31cosecθ 20 0 − + = 2 12cosec θ 16 cosec θ 15cosecθ 20 0 − − + = 4cosecθ(3cosecθ 4) 5(3cosecθ 4) 0 − − − = (3cosecθ 4)(4cosecθ 5) 0 − − = 4 cosec θ 3 ⇒ = and 5 cosec θ 4 = 3 sin θ 4 = and 4 sin θ 5 = 43 sin θ , 54 ⇒ 188. Find the smallest positive angle which satisfies the given trigonometric equation. 2 2sin x + 3cosx + 1 = 0 (a) 6 π (b) 3 π (c) 2 3 π (d) 5 6 π SSC CHSL –19/10/2020 (Shift-I)

Trigonometry 138 . Ans. (d) : 2 2sin x 3 cos x 1 0 + + = ( ) 2 2 1–cos 3 cos 1 0 x x + + = 2 2 cos 3 cos 3 0 x x − − = 2 2 cos 2 3 cos 3 cos 3 0 x x x − + − = 2 cos (cos 3) 3(cos 3) 0 x x x − + − = (cos 3)(2cos 3) 0 x x − + = 3 cos cos150 2 x − = = ° ∴ 5 6 x π = 189. If sinθ + cosecθ = 2, then the value of sin 2 θ + cosec 2 θ is: (a) 1 (b) 2 (c) 8 (d) 4 SSC CHSL –16/10/2020 (Shift-II) Ans. (b) : sin cosec 2 θ+ θ= Q or 1 1 sin 2 [ cosec ] sin sin θ+ = θ= θ θ Q or 2 2 1 sin 2 sin   θ+ =   θ   [By squaring both sides] or 2 2 1 sin 2 4 sin θ+ + = θ or 2 2 1 sin 2 sin θ+ = θ or 2 2 sin cos ec 2 θ+ θ= 190. cosx 1 + sinx + 1 + sinx cosx is equal to: (a) 2cosx (b) 2cosecx (c) 2sinx (d) 2secx SSC CHSL –15/10/2020 (Shift-I) Ans. (d) : cos x 1 sin x 1 sin x cos x + + + ( ) ( ) ( ) 2 2 2 2 cos x sin x 1 2 sin x cos x 1 sin x 2 sin x cosx 1 sin x cosx 1 sin x + + + + + + = = + + ( ) ( ) ( ) 21 sin x 2 2sinx 2 2secx cosx 1 sin x cosx 1 sin x cos x + + = = = = + + 191. If tanx = m n and 0 o ≤ x ≤ 90 o , then the value of (sinx + cosx) is : (a) 2 2 1 m n − (b) 2 2 m+n m n + (c) 2 2 m n + (d) 2 2 1 m n + SSC CHSL –13/10/2020 (Shift-III) Ans. (b) : From question, m Height tanx = n Base = Hypotenuse · ( ) ( ) 2 2 AB BC + = 2 2 m n + Hence sin x + cos x = Height Base Base Height + = 2 2 2 2 m n m n m n + + + = 2 2 m n m n + + 192. cot x 1 + cosec x + 1 + cosec x cot x is equal to: (a) 2sin x (b) 2cosec x (c) 2sec x (d) 2cos x SSC CHSL –13/10/2020 (Shift-III) Ans. (c) : Given- cot x 1 cosec x 1 + cosec x cot x + + = ( ) 2 2 cot x 1 cosec x 2cosec x cot x 1 + cosec x + + + = ( ) 2 2 cosec x cosec x 2cosec x cot x 1 + cosec x + + = ( ) 2 2cosec x 2cosec x cot x 1 cosec x + + = ( ) ( ) 2cosec x 1 + cosec x cot x 1 + cosec x = sin x 2 sin x cos x × = 2 sec x 193. If cot A = k, then sin A is equal to: (asume that A is an acute angle) (a) 1 k (b) 1 k − (c) 2 1 1 k + (d) 2 2 k 1 k + SSC CHSL –12/10/2020 (Shift-I) Ans. (c) : Given-

Trigonometry 139 . Base cot A Height = Q ∴ cot A = K 1 Base · K, Height · 1 Base · 2 K 1 + Base sin A Height = Q sin A = 2 1 k 1 + 194. If the value of sec B + tan B = r, then the value of sec B – tan B is equal to: (a) 1/r (b) 0 (c) r 2 (d) –r SSC CHSL –20/10/2020 (Shift-III) Ans : (a) sec B + tan B = r (Given) {Qsec 2 θ-tan 2 θ = 1} (sec B + tan B) (sec B – tan B) = 1 (sec B – tan B) = 1 r 195. If ( ) 3 tan , 0º 90º 5 θ= <θ< θ= <θ< θ= <θ< θ= <θ< then sinθ.cosθ is equal to : (a) 15 34 (b) 4 34 1 (c) 17 (d) 16 34 SSC CHSL 11/07/2019 (Shift-III) Ans. (a) : From Pythagoras theorem AC 2 = BC 2 + AB 2 AC 2 = 3 2 + 5 2 AC 2 = 34 AC 34 = 3 sin , 34 θ= 5 cos 34 θ= 3 5 sin .cos 34 34 ∴ θ θ= × 15 34 = 196. If cos cos x sin cos sin cos sin cos α β α β α β α β + = + + = + + = + + = + α+ β β− α α− β α+ β β− α α− β α+ β β− α α− β α+ β β− α α− β cos sin cos β β+ α β+ α β+ α β+ α then x is equal to : (a) sinβ (b) sinα (c) cosβ (d) cosα SSC CHSL 10/07/2019 (Shift-I) Ans. (d) cos cos x cos sin cos sin cos sin cos sin cos α β β + = + α+ β β− α α− β β+ α Let ( ) 0 0 180 180 α+β= ⇒β= −α cos cos cos cos ∴ α=− β⇒ β=− α ( ) ( ) 0 0 cos cos sin cos 180 sin 180 cos α β ∴ + α+ −α −α − α ( ) ( ) 0 0 x cos sin cos 180 sin 180 cos β = + α− −α −α + α cos cos sin cos sin cos α α   ⇒ −   α− α α− α   ( ) ( ) x cos sin cos sin cos α = − α+ α α+ α ( ) x cos 0 sin cos − α = α+ α x cos 0 − α= x cos = α 197. If 1 sin x a tan x 1 sin x − = − = − = − = − + then a is equal : (a) cosx (b) sinx (c) secx (d) cosecx SSC CHSL 10/07/2019 (Shift-III) Ans. (c): { } 2 2 1 sin x a tan x 1 sin x cos x 1 sin x − = − − = + Q ( ) 1 sin x a tan x cos x − = − sec x tan x a tan x − = − a sec x = 198. If cos 4x θ= θ= θ= θ= and 4 sin (x 0) x θ= ≠ θ= ≠ θ= ≠ θ= ≠ then find the value of 2 2 1 x x         +                 : (a) 1 2 (b) 1 16 (c) 1 4 (d) 1 3 SSC CHSL 09/07/2019 (Shift-I) Ans. (b) : Given, 4 cos 4x, sin x θ= θ= cos x , 4 θ = 1 sin x 4 θ = 2 2 2 2 1 cos sin x 16 16 x θ θ + = + = 2 2 1 (cos sin ) 16 θ+ θ = 1 16 199. If θ is an acute angle and given 5sinθ + 12cosθ = 13, then find the value of tanθ : (a) 12 13 (b) 5 13 (c) 5 12 (d) 13 12 SSC CHSL 09/07/2019 (Shift-III)

Trigonometry 140 . Ans. (c) : 5sin 12cos 13 θ+ θ= 5 12 sin cos 1 ...........(i) 13 13 θ+ θ= Q 2 2 sin cos 1 θ+ θ= ∴ sin .sin cos .cos 1 ............(ii) θ θ+ θ θ= By comparing the equation (i) and (ii) 5 12 sin , cos 13 13 θ= θ= 5 sin 5 13 13 12 cos 13 12 13 θ = = × θ 5 tan 12 θ= 200. If sinθ = 3x and ( ) 3 cos θ ,x 0 x = ≠ = ≠ = ≠ = ≠ then value of 2 2 1 6x x         +                 : (a) 1 4 (b) 1 2 (c) 1 3 (d) 2 3 SSC CHSL 08/07/2019 (Shift-II) Ans. (d) : 2 2 sin θ 3x, sin θ 9x ........... (i) = = 2 2 3 9 cos θ , cos θ ...........(ii) x x = = By adding the equation (i) and (ii) 2 2 2 2 9 sin θ cos θ 9x x + = + 2 2 1 1 9x x   = +     2 2 1 1 x 9 x = + Multiplying by 6 in both side. 2 2 1 6 6x 9 x   + =     2 3 = (III) Problems based on Trigonometric Functions 201. If cosecθ = 17 15 , then the value of cosθ will be: (a) 1 (b) 8 17 (c) 15 17 (d) 7 17 SSC CHSL (Tier-I) 14/08/2023 (Shift-IV) Ans. (b) : Given that, cosecθ = 17 15 ∵ sinθ = 1 15 cosec 17 = θ Now, cos θ = 2 1 sin − θ ⇒ cos θ = 2 15 1 17   −     ⇒ cos θ = 289 225 289 − ⇒ cos θ = 64 8 289 17 = 202. If sinθ + cosθ = 1 29 , then find the value of sinθ + cosθ sinθ – cosθ . (a) 43 29 (b) 1 41 (c) 1 43 (d) 41 29 SSC CHSL (Tier-I) 03/08/2023 (Shift-II) Ans. (b): Given that sin θ + cos θ = 1 29 ⇒ (sin θ + cos θ ) 2 = 2 1 29       ⇒ sin 2 θ + cos 2 θ + 2sin θ .cos θ = 1 1 29 29 × ⇒ 1 + sin2 θ = 1 841 (sin 2 θ + cos 2 θ = 1 & sin2 θ = 2sin θ .cos θ ) ⇒ sin2θ = 1 1 841 − ⇒ sin2θ = 840 841 − Now, (sin θ – cos θ ) 2 = (sin θ + cos θ ) 2 – 4sin θ cos θ ⇒ (sin θ – cos θ ) 2 = 2 1 2 sin 2 29   − θ     ⇒ (sin θ – cos θ ) 2 = ( ) 2 840 1 841 841 − − ⇒ (sin θ – cos θ ) 2 = 1681 841 ⇒ sin θ – cos θ = 41 29 So, sin cos 1/29 1 sin cos 41/ 29 41 θ+ θ = = θ− θ 203. If secθ + θ + θ + θ + tanθ = θ = θ = θ = 3 , then find the positive value of (sinθ + θ + θ + θ + cosθ). θ). θ). θ). (a) 3 2 (b) 2 3 (c) 2 (1 3) + (d) (1 3) 2 + SSC CHSL (Tier-I) 04/08/2023 (Shift-III)

Trigonometry 141 . Ans. (d) : Given that secθ + tanθ = 3 ⇒ 1 sin 3 cos + θ = θ ⇒ (1+sinθ) = ( 3 cosθ) ⇒ (1+sinθ) 2 = ( 3 cos θ) 2 ⇒ 1+sin 2 θ + 2sinθ = 3cos 2 θ ⇒ 1+sin 2 θ + 2sinθ = 3(1–sin 2 θ) ⇒ 4sin 2 θ + 2sinθ – 2 = 0 ⇒ 2sin 2 θ + sinθ – 1 = 0 ⇒ (sinθ + 1) (2sinθ – 1) = 0 ⇒ sinθ = – 1, 1/2 ⇒ sinθ = 1/2 (Taking positive value) sinθ = sin30° ⇒ θ = 30º ∴ sinθ + cosθ = sin30º + cos30º = 1 3 2 2 + ⇒ (sinθ + cosθ) = ( 3 1) 2 + 204. If sinA + sin 2 A = 1, then the expression (cos 2 A + cos 4 A) is _____ (a) 1 (b) 1 2 (c) 3 2 (d) 2 SSC CHSL (Tier-I) 02/08/2023 (Shift-I) Ans. (a) : cos 2 A + cos 4 A = ? sin A + sin 2 A = 1, 2 sin A 1 sin A = − 2 sin A cos A = On squaring both the sides 2 4 1 cos A cos A − = 2 4 1 cos A cos A − = 2 2 sin A 1 cos A = − Q 2 4 1 cos A cos A = + 205. Simplify the given equation 3 cot A – 1 cotA – 1 (a) cot 2 A + cosecA (b) cosec 2 A + cotA (c) cosec 2 A + cot 2 A (d) cot 2 A - cosecA SSC CHSL (Tier-I) 08/08/2023 (Shift-II) Ans. (b) : 3 2 cot A 1 (cot A 1)(cot A 1 cot A) cot A 1 (cot A 1) − − + + = − − 2 cot A 1 cot A = + + 2 cosec A cot A = + 2 2 [ cosec A 1 cot A] = + Q 206. If 4sin 2 A – 3 = 0 and 0 ≤ A ≤ 90º, then find the value of (3sinA – 4sin 3 A). (a) 3 2 (b) 1 2 (c) 1 (d) 0 SSC CHSL (Tier-I) 10/08/2023 (Shift-I) Ans. (d) : 4 sin 2 A –3 = 0 2 3 sin A 4 ⇒ = 3 si n A 2 = sinA = sin60° 0 0 A 60 [ 0 A 90 ] = ≤ ≤ Q ∴ 3sinA – 4sin 3 A = 3sin60 0 – 4sin 3 60 0 3 3 3 3 4 2 2   × = − ×     33 33 0 2 2 = − = 207. If sinA = 2 5 where A is an acute angle, then the value of 5sinA + 2cosecA 21secA is : (a) 4 5 (b) 5 4 (c) 5 7 (d) 7 5 SSC CHSL (Tier-I) 10/08/2023 (Shift-I) Ans. (d) : 2 P sin A 5 H = = In ΔABC by Pythagoras Theorem AB 25 4 21 = − = 5 5 cosecA , sec A 2 21 = = So, 2 5 5 2 5sin A 2cosecA 5 2 5 21 sec A 21 21 × + × + = × 7 5 = 208. If 21cosA + 3sinA 3cosA + 4sinA = 2, then find the value of cotA. (a) 1 3 (b) 11 9 (c) 11 10 (d) 9 10 SSC CGL (Tier-I) 26/07/2023 (Shift-II) Ans. (a) : 21cos A 3sin A 2 cot A ? 3cosA 4 sin A + = = + 21cos A 3sin A 6cosA 8sin A + = + 15cosA 5sin A = cos A 5 1 cot A sin A 15 3 = ⇒ = 209. If cosx + sinx = 2cosx , then the value of (cosx – sinx) 2 + (cosx + sinx) 2 is : (a) 0 (b) 1 (c) 1 2 (d) 2 SSC CGL (Tier-I) 19/07/2023 (Shift-IV)

Trigonometry 142 . Ans. (d) : Given that cosx + sinx = 2cosx (cosx – sinx) 2 + (cosx + sinx) 2 = (cos 2 x + sin 2 x) – 2cosx sinx + (cos 2 x + sin 2 x) + 2cosx sinx = 1 + 1 = 2 210. If sinθ – cosθ = 0°, then the value of (sin 3 θ – cos 3 θ) is : (a) 2 (b) 1 (c) 0 (d) 1 2 SSC CGL (Tier-I) 19/07/2023 (Shift-IV) Ans. (c) : sin cos 0 θ− θ= then (sin 3 θ – cos 3 θ) = ? sin cos 0 θ− θ= sin cos θ= θ 0 sin 1 tan 1 45 cos θ = ⇒ θ= ⇒θ= θ ∴ sin 3 θ – cos 3 θ = 3 3 3 3 1 1 sin 45 cos 45 0 2 2     °− °= − =         211. If {(3sinθ – cosθ)/(cosθ + sinθ)} = 1, then find the value of cotθ. (a) 3 (b) 1 (c) 2 (d) 0 SSC CGL (Tier-I) 14/07/2023 (Shift-I) Ans. (b) : 3sin cos 1 cos sin θ− θ = θ+ θ , then cot ? θ= 3sin cos cos sin θ− θ= θ+ θ cos cos 3sin sin θ+ θ= θ− θ 2 cos 2sin θ= θ cos 2 cot 1 sin 2 θ = ⇒ θ= θ 212. What will be the value of the following? ( )( )       3 3 sinAcosA 1 + cotA + tanA sinA - cosA sin A - cos A (a) –1 (b) 0 (c) 2 (d) 1 SSC CHSL Tier- II 26/06/2023 Ans.(d) : ( )( ) 3 3 sin A.cos A 1 cot A tan A sin A cos A sin A cos A   + + −   −   ( )( ) ( ) ( ) 2 2 sin A.cos A 1 cot A tan A sin A cos A sin A cos A sin A cos A sin A.cos A     + + −   − + +   2 2 cos A sin A sin A.cos A 1 sin A cos A sin A cos A sin A.cos A    = + +    + +    2 2 2 2 sin A.cos A cos A sin A sin A.cos A sin A.cos A sin A cos A sin A.cos A   + +   =    + +     = 1 213. If (1 + tanθ) = θ) = θ) = θ) = 3 , then 3 cotθ θ θ θ – 1 is ____. (a) 2 3 1 2 − (b) 2 3 1 2 + (c) 3 1 2 − (d) 3 1 2 + SSC CHSL –03/06/2022 (Shift-II) Ans. (d) : Given that, 1 + tan θ = 3 tan θ = 3 – 1 1 cot 3 1 θ= − So, 1 3 cot 1 3 1 3 1   θ− = −   −   3 ( 3 1) 3 1 − − = − 3 3 1 3 1 − + = − 1 3 1 = − 1 ( 3 1) 3 1 ( 3 1) × + = −× + 3 1 3 1 3 1 2 + + = = − 214. If 2sin 2 θ + 3cosθ = 3, 0 0 < θ < 90 0 , then find the value of (sec 2 θ + cot 2 θ). (a) 2 3 3 (b) 1 4 3 (c) 1 3 3 (d) 1 4 2 SSC CGL 18/04/2022 (Shift-III) Ans. (b) : 2sin 2 θ + 3cosθ = 3, 0 0 < θ < 90 0 2 (1–cos 2 θ) + 3 cosθ = 3 {Q sin 2 θ = 1–cos 2 θ) 2 – 2 cos 2 θ + 3 cosθ = 3 2cos 2 θ – 3 cosθ + 1 = 0 2cos 2 θ – 2 cosθ – cosθ + 1 = 0 (2 cosθ –1) (cosθ –1) = 0 2cosθ –1 = 0 1 cos 2 θ= cosθ = cos60 0 0 60 θ= So, sec 2 θ + cot 2 θ = sec 2 60 0 + cot 2 60 0 () 2 2 1 2 3   = +     1 4 3 = + 1 4 3 = 215. If 6 tanA(tanA + 1) = 5 – tanA, Given that 0 < A < 2 π what is the value of (sinA + cosA). (a) 5 3 (b) 35 (c) 5 3 (d) 3 5 SSC CGL 20/04/2022 (Shift-I) Ans. (d) : 6tanA (tanA+1) = 5 – tanA 6tan 2 A + 6 tanA + tanA = 5 6tan 2 A + 7 tanA – 5 = 0

Trigonometry 143 . By Sridharacharya Rule, 6tan 2 A + 10tanA – 3 tan A – 5 = 0 2tan(3tanA + 5) – 1 (3tan A + 5) = 0 (2tan A – 1) (3tan A + 5) = 0 tan A = 1 2 Sec 2 A – tan 2 A = 1 2 2 1 sec A 1 2   = +     5 sec A 2 = 2 cos A 5 = sin 2 A + cos 2 A = 1 2 4 sin A 1 5 + = 1 sin A 5 = According to the question, 1 2 sin A cos A 5 5 + = + 3 sin A cos A 5 + = 216. If 7sin 2 θ + 4cos 2 θ = 5 and θ is in the first quadrant, then the value of 3secθ + tanθ 2cotθ - 3cosθ wll be : (a) ( ) 2 2 1 − (b) ( ) 21 2 + (c) 3 2 (d) 4 2 SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (b) : 7sin 2 θ + 4cos 2 θ = 5 3sin 2 θ + 4 sin 2 θ + 4 cos 2 θ = 5 3 sin 2 θ + 4 = 5 (Q sin 2 θ + cos 2 θ = 1) 3sin 2 θ = 1 sinθ = 1 3 = P H By Pythagoras theorem, H 2 = P 2 + B 2 3 – 1 = B 2 B = 2 ∴ 3 1 3 3 sec tan 2 2 2 cot 3 cos 2 2 2 3 1 3 × + θ+ θ = θ− θ × − × 4 2 2 2 2 2 2 2 = = − − ( ) ( ) ( ) 2 22 2 2 2 2 2 + = − + ( ) 2 22 2 2 + = ( ) 22 2 = + ( ) 21 2 = + 217. If cosec θ + cot θ =7 cosec θ - cotθ , then the value of 2 2 4sin θ-1 4sin θ + 5 is : (a) 1 3 (b) 1 9 − (c) 1 9 (d) 1 3 − SSC CPO SI 11/11/2022 (Shift III rd ) Ans. (c) : cos ec cot 7 cos ec cot 1 θ+ θ = θ− θ 7cosecθ – 7 cotθ = cosecθ + cotθ 6 cosecθ = 8cotθ 1 cos 6 8 sin sin θ × = × θ θ 6 = 8 cosθ 3 cos 4 θ= 2 9 7 sin 1 cos 1 16 4 θ= − θ= − = So, 2 2 4sin 1 4sin 5 θ− θ+ 7 4 1 16 7 4 5 16 × − = × + 3 1 4 27 9 4 = = 218. If sec – tan 1 = sec + tan 7 θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ , and θ is in first quadrant, then find the value of 2 2 cosecθ + cot θ cosecθ - cot θ : (a) 19 5 (b) 37 19 (c) 22 3 (d) 37 12 SSC CPO SI 10/11/2022 (Shift II nd ) Ans. (a) : sec tan 1 sec tan 7 θ− θ = θ+ θ 1 sin 1 cos cos 1 sin 7 cos cos θ − θ θ = θ + θ θ 1 sin 1 cos 1 sin 7 cos − θ θ = + θ θ 1 sin 1 1 sin 7 − θ = = + θ

Trigonometry 144 . 7–7 sinθ = 1 + sinθ 8sinθ = 6 3 sin 4 θ= 2 3 7 cos 1 sin 1 4 4 2   θ= − θ= − =     4 cos 7/4 7 cosec , cot 3 sin 3/4 3 θ θ= θ= = = θ ∴ 2 2 2 2 4 7 3 3 cos ec cot cos ec cot 4 7 3 3   +     θ+ θ   = θ− θ   −       4 7 3 9 4 7 3 9 + = − 12 7 9 12 7 9 + = − 19 5 = 219. If cos 2 θ – sin 2 θ = tan 2 φ, then which of the following is true? (a) cos 2 φ – sin 2 φ = cot 2 θ (b) cos 2 φ – sin 2 φ = tan 2 θ (c) cosθ cosφ = 2 (d) cosθ cosφ = 1 SSC CPO-SI 13/12/2019 (Shift-II) Ans. (b) cos 2 θ – sin 2 θ = tan 2 φ cos2θ = tan 2 φ 2 2 1 tan 1 tan − θ + θ = tan 2 φ From the componendo and dividendo method 2 2 2 2 1 tan 1 tan 1 tan 1 tan − θ+ + θ − θ− − θ = 2 2 tan 1 tan 1 φ+ θ− 2 2 2 tan − θ = 2 2 1 tan (1 tan ) + φ − − φ = 1 cos 2 − φ tan 2 θ = cos2φ tan 2 θ = cos 2 φ – sin 2 φ 220. If 2 tan = 13 a , then the value of 2 2 2 2 cosec α + 2sec α cosec α - 3sec α is: (a) 21 (b) 14 (c) 32 (d) 16 SSC CHSL –18/03/2020 (Shift-I) Ans. (a) Q tan α = 2 13 ∴ cot a = 13 2 then, 2 2 2 2 2 2 2 2 cosec 2sec 1 cot 2 2 tan cosec 3sec 1 cot 3 3 tan α+ α + α+ + α = α− α + α− − α = 2 2 2 2 13 4 3 2 3 cot 2 tan 4 13 13 4 2 cot 3 tan 2 3 4 13 + + × + α+ α = − + α− α − + − × = 3 52 169 32 156 201 2 52 169 48 104 121 × + + + = −× + − − + 357 = 21 17 = 221. If sin θ + cosec θ = 7, then what is the value of sin 3 θ + cosec 3 θ? (a) 350 (b) 382 (c) 322 (d) 367 SSC CHSL 06/082021 (Shift-II) Ans. (c) : sin θ + cosec θ = 7 → sin 2 θ + cosec 2 θ+2=49 sin 3 θ + cosec 3 θ = (sinθ + cosθ) [sin 2 θ+cosec 2 θ–1] = 7 × [47–1] = 7 × 46 = 322 222. If (1+cot 2 θ) + {1+(cot 2 θ) –1 } is equal to k, then k = ? (a) sin θ cos θ (b) cosec θ sec θ (c) sin θ sec θ (d) cosec θ cos θ SSC CHSL 10/08/2021 (Shift-III) Ans. (b) : ( ) ( ) { } 1 2 2 1 cot 1 cot k − + θ+ + θ = ( ) ( ) 2 2 1 cot 1 tan k + θ+ + θ= 1 tan cot   θ=   θ   Q cosec 2 θ + sec 2 θ= k 2 2 2 2 2 2 sin cos 1 k k sin cos co sec sec θ+ θ = ⇒ = θ θ θ θ k cosec sec ∴ = θ θ 223. If tan θ + cot θ = 3, then what will be the value of tan 2 θ + cot 2 θ? (a) 11 (b) 1 (c) –1 (d) 7 SSC CHSL 12/08/2021 (Shift-I) Ans. (d) : tan θ + cot θ = 3 square both sides, tan 2 θ + cot 2 θ +2tan θ cot θ = 9 tan 2 θ + cot 2 θ = 9–2 = 7 ∴ tan 2 θ + cot 2 θ = 7 224. If sinx – cosx = 0, then the value of (sin 3 x – cos 3 x) is: (a) 2 (b) 4 (c) 0 (d) 1 SSC CHSL –13/10/2020 (Shift-III) Ans. (c) : sinx – cosx = 0 Q ∴(sinx – cosx) 3 = sin 3 x – cos 3 x–3sinx cosx (sinx – cosx) 0 = sin 3 x – cos 3 x – 3 sinx cosx ×0 or sin 3 x – cos 3 x = 0

Trigonometry 145 . 225. If tan 4 x – tan 2 x = 1, then the value of sin 4 x + sin 2 x is: (a) 3 4 (b) 3 2 (c) 1 2 (d) 1 SSC CHSL –14/10/2020 (Shift-II) Ans. (d) : tan 4 x – tan 2 x = 1 tan 4 x = 1 + tan 2 x tan 4 x = sec 2 x 4 4 2 sin x 1 cos x cos x = sin 4 x = cos 2 x sin 4 x + sin 2 x = cos 2 x + sin 2 x = 1 226. If cos 2 θ + cos 4 θ = 1, then the value of sinθ + sin 2 θ is: (a) 1 2 (b) 0 (c) 1 (d) 2 SSC CHSL –16/10/2020 (Shift-I) Ans. (c) : cos 2 θ + cos 4 θ = 1 or cos 4 θ = 1 – cos 2 θ or cos 4 θ = sin 2 θ or cos 2 θ = sinθ ∴ sinθ + sin 2 θ = cos 2 θ + sin 2 θ = 1 227. If cot 6 θ= θ= θ= θ= then the value of 2 2 2 2 cosec + sec cosec - sec θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ is : (a) 49 36 (b) 48 35 (c) 7 5 (d) 43 36 SSC CHSL 03/07/2019 (Shift-I) Ans. (c) : 6 cot 1 θ= From the Pythagoras theoram, ( ) () 2 2 2 AC 6 1 = + 2 AC 6 1 = + AC 7 = As per question, ( ) ( ) 2 2 2 2 2 2 2 2 7 7 6 cosec sec cosec sec 7 7 6   +     θ+ θ   = θ− θ   −       7 7 49 6 6 7 6 35 7 6 + = = × − 49 7 35 5 = = 228. The value of 4 4 2 2 sin θ + cos θ + 2sin θcos θ is: (a) 2 (b) 0 (c) 1 (d) 4 SSC CHSL –19/10/2020 (Shift-III) Ans. (c) : 4 4 2 2 sin θ + cos θ + 2sin θcos θ = ( ) 2 2 2 sin cos θ+ θ = (1) 2 ⇒ 1 229. If 4cos 2 θ – 3sin 2 θ + 2 = 0 then the value of tanθ is (where 0 0 ≤ θ < 90 O ) (a) 6 (b) 1 (c) 1 3 (d) 2 SSC CHSL –12/10/2020 (Shift-II) Ans. (a) : 4cos 2 θ – 3sin 2 θ + 2 = 0 4 cos 2 θ – 3(1 – cos 2 θ) = – 2 4 cos 2 θ – 3 + 3 cos 2 θ = – 2 7 cos 2 θ = 1 cos θ 1 7 = Q 6 tan 1 θ= 6 = 230. The value of 2 2 2 2 tan + cot - sec cosec φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ is equal to : (a) –2 (b) 1 (c) 0 (d) –1 SSC CGL (Tier-II) 12-09-2019 Ans. (a) tan 2 φ + cot 2 φ – sec 2 φ cosec 2 φ 2 2 2 2 2 2 sin cos 1 cos sin sin cos φ φ = + − φ φ φ φ ( ) 2 2 2 2 2 4 4 2 2 2 2 sin cos 2sin cos 1 sin cos 1 cos sin cos sin φ+ φ − φ φ− φ+ φ− = = φ φ φ φ 2 2 2 2 2 2 2 2 1 2sin cos 1 2sin cos 2 sin cos sin cos − φ φ− − φ φ = = =− φ φ φ φ 231. – – 2 1 tan 1 ? 1 cot θ         + = + = + = + =         θ         (a) cosec 2 θ (b) sec 2 θ (c) cos 2 θ (d) sin 2 θ SSC CGL (Tier-II) 11-9-2019

Trigonometry 146 . Ans. (b) : 2 1 tan 1 1 cot − θ  +   − θ   2 1 tan 1 1 1 tan     − θ = +     −   θ   ( ) ( ) 2 2 1 tan 1 tan 1 tan 1 tan     − θ = + =− θ +   θ−     θ   = tan 2 θ + 1 = sec 2 θ 232. (secφ– tanφ) 2 (1+sinφ) 2 ÷ sin 2 φ = ? (a) secφ (b) cot 2 φ (c) costφ (d) cosφ SSC CGL (Tier-II) 12-09-2019 Ans. (b) : (secφ– tanφ) 2 (1+sinφ) 2 ÷ sin 2 φ ( ) 2 2 2 1 sin 1 sin sin cos cos   φ = − + φ ÷ φ   φ φ   ( ) ( ) 2 2 2 2 1 sin 1 sin sin cos − φ × + φ = ÷ φ φ ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 sin cos cos sin cos cos sin sin − φ φ φ = ÷ φ= = φ φ× φ φ = cot 2 φ 233. The value of             2 sinA 1 - cosA cot A + ÷ +1 1 - cosA sinA 1 + cosecA is : (a) 2 (b) 3 2 (c) 1 (d) 1 2 SSC CGL (Tier-II) 12-09-2019 Ans. (a) : 2 sin A 1 cos A cot A 1 1 cos A sin A 1 cos ecA   −   + ÷ +     − +     ( ) 2 2 2 2 sin A 1 cos A 2cosA cos A 1 sin A 1 cos A sin A sinA   + + − = ÷ +   − +   ( ) ( ) ( ) ( ) 2 21 cos A 1 sin A sin A 1 cosA sin A sin A − + = ÷ − + ( ) sinA 1 sin A 2 2 sin A 1 sin A + = × = + 234. The value of ( ) 2 2 2 2 2 2 sec θ cosec θ + - sec θ + cosec θ cosec θ sec θ is : (a) 1 (b) 0 (c) –2 (d) 2 SSC CGL (Tier-II) 13-09-2019 Ans. (c) : ( ) 2 2 2 2 2 2 sec θ cosec θ + sec θ + cosec θ cosec θ sec θ - 2 2 2 2 2 2 sin cos 1 1 cos sin cos sin θ θ   = + − +   θ θ θ θ   ( ) 4 4 2 2 2 2 sin cos sin cos cos .sin θ+ θ− θ+ θ = θ θ ( ) 2 2 2 2 2 2 2 sin cos 2sin .cos 1 cos .sin θ+ θ − θ θ− = θ θ 2 2 2 2 1 2sin .cos 1 cos .sin − θ θ− = θ θ = –2 235. The value of ( ) ( ) 6 6 4 4 4 4 2 2 sin θ + cos θ - 3 sin θ + cos θ cos θ - sin θ - 2cos θ is: (a) 1 (b) 2 (c) –2 (d) –1 SSC CGL (Tier-II) 13-09-2019 Ans. (a) : ( ) ( ) 6 6 4 4 4 4 2 2 sin cos 3 sin cos cos sin 2cos θ+ θ− θ+ θ θ− θ− θ ( ) ( )( ) 2 2 4 4 2 2 4 4 2 2 2 2 2 2 sin cos sin cos sin .cos 3 sin cos cos sin cos sin 2cos    θ+ θ θ+ θ− θ θ− θ+ θ    = θ+ θ θ− θ− θ ( ) ( ) 4 4 2 2 4 4 2 2 2 2 sin cos sin .cos 3 sin cos cos sin 2 cos θ+ θ− θ θ− θ+ θ = θ− θ− θ ( ) 4 4 2 2 2 2 sin cos 2sin .cos cos sin − θ− θ− θ θ = − θ+ θ ( ) ( ) 2 2 2 2 2 sin cos cos sin − θ+ θ = − θ+ θ 1 1 1 = = 236. The value of 2 2 2 3(1 - 2sin x) cos x-sin x is: (a) 1 (b) 4 (c) 3 (d) 2 SSC CGL (Tier-I) – 09/03/2020 (Shift-III) Ans. (c) : ( ) 2 2 2 31 2sin x cos x sin x − − ( ) ( ) 2 2 2 2 31 2sin x cos x 1 sin x 1 2sin x − = = − − Q = 3 237. The expression 3sec 2 θ tan 2 θ + tan 6 θ – sec 6 θ is equal to: (a) –1 (b) 1 (c) 2 (d) –2 SSC CGL (Tier-I) – 04/03/2020 (Shift-II) Ans. (a) : 3sec 2 θ.tan 2 θ + tan 6 θ – sec 6 θ = 3sec 2 θ.tan 2 θ – (sec 6 θ–tan 6 θ) = 3sec 2 θ.tan 2 θ – (sec 2 θ – tan 2 θ) (sec 4 θ+tan 4 θ+sec 2 θ.tan 2 θ) = 3sec 2 θ.tan 2 θ – 1 × [(sec 2 θ – tan 2 θ) 2 + 3sec 2 θ.tan 2 θ] = 3sec 2 θ.tan 2 θ – 1 – 3sec 2 θ.tan 2 θ = –1

Trigonometry 147 . 238. The value of 6 6 2 2 4 4 2 sec tan 3sec tan 1 cos sin 2sin 2 θ− θ− θ θ+ θ− θ+ θ+ is: (a) 1 (b) 1 2 (c) 2 3 (d) 3 4 SSC CGL (Tier-I) – 05/03/2020 (Shift-I) Ans. (c) : 6 6 2 2 4 4 2 sec θ - tan θ - 3sec θtan θ +1 cos θ - sin θ + 2sin θ + 2 = 2 2 4 4 2 2 2 2 (sec tan )(sec tan sec tan ) 3 sec . tan 1 2 2 2 2 2 (cos sin ) (cos sin ) 2 sin 2 θ− θ θ+ θ+ θ θ− θ θ+ θ+ θ θ− θ+ θ+ = 2 2 2 2 2 2 2 2 2 2 {(sec tan ) 3sec .tan } 3sec .tan 1 1 1 sin sin 2sin 2 θ− θ + θ θ− θ θ+ × − θ− θ+ θ+ 2 2 2 2 sec tan 1 cos 1 sin   θ− θ=     θ= − θ   Q = 1 1 2 3 3 + = 239. The value of       2 2 2 (1 - secA) + (1 + secA) 4 1+sec A is: (a) 2 (b) 8 (c) 1 (d) 4 SSC CGL (Tier-I)– 06/03/2020 (Shift-III) Ans. (b) : 2 2 2 (1 sec A) (1 sec A) 4 1 sec A   − + +   +   = 2 2 2(1 sec A) 4 1 sec A   +   +   2 2 2 2 (a b) (a b) 2(a b)   + + − = +   Q = 8 240. If cosθ + sec θ = 2, then (cos 117 θ + sec 117 θ) equal to : (a) 234 (b) 2 117 (c) 2 (d) 117 SSC CGL (TIER-I) – 19.06.2019 (Shift-III) Ans. (c) cosθ + sec θ = 2 Let θ = 0 0 cos0 0 + sec0 0 = 2 1 + 1 = 2 2 = 2 ∴ cos 117 θ+sec 117 θ=cos 117 0 0 +sec 117 0 0 =(1) 117 +(1) 117 = 2 241. The value of 2 2 2 2 sec cosec tan sin θ+ θ× θ− θ θ+ θ× θ− θ θ+ θ× θ− θ θ+ θ× θ− θ is equal to : (a) sinθ cos 2 θ (b) sinθ sec 2 θ (c) cosecθ cos 2 θ (d) cosecθ sec 2 θ SSC CGL (TIER-I) – 06.06.2019 (Shift-I) Ans. (b) : 2 2 2 sec cosec tan sin 2 θ+ θ× θ− θ 2 2 2 2 1 1 sin sin cos sin cos 2 θ = + × − θ θ θ θ 2 2 2 2 2 sin cos sin 1 cos .sin cos 2 θ+ θ 1   = × θ −   θ θ θ   2 2 2 1 sin .sin cos sin cos θ θ = × θ. θ θ 2 1 sin cos .sin cos θ = × θ θ θ 2 sin .sec = θ θ 242. If ( ) ( ) (( ( ) ) ) 1 2 2 1 tan 1 tan k, − + θ+ + θ = + θ+ + θ = + θ+ + θ = + θ+ + θ = then k ? = (a) cosecθ secθ (b) sinθ cosθ (c) cosecθ cosθ (d) sinθ secθ SSC CGL (TIER-I)– 06.06.2019 (Shift-III) Ans. (a): ( ) ( ) 1 2 2 1 tan 1 tan k −   + θ+ + θ =     2 2 sec 1 cot k   θ+ + θ =   2 2 sec cosec k θ+ θ= 2 2 1 1 k cos sin + = θ θ 2 2 2 2 2 2 sin cos 1 k sin .cos sin .cos θ+ θ = = θ θ θ θ 2 2 cosec .sec = θ θ k cosec .sec = θ θ 243. The expression (cos 6 θ+sin 6 θ–1)(tan 2 θ+cot 2 θ+2) +3 is equal to: (a) 0 (b) 1 (c) 2 (d) –1 SSC CGL (Tier-I) 13/04/2022 (Shift-II) Ans : (a) (cos 6 θ + sin 6 θ-1) (tan 2 θ + cot 2 θ + 2) + 3 = ? = [(cos 2 θ + sin 2 θ) ( cos 4 θ + sin 4 θ – cos 2 θ – sin 2 θ)–1](tanθ+cotθ) 2 + 3 ( ) 2 2 2 2 2 2 2 2 2 2 sin cos 1 cos sin 2cos .sin cos .sin 1 3 sin .cos θ+ θ     θ+ θ − θ θ− θ θ− +     θ θ   ( ) 2 2 2 2 1 1 3cos .sin 1 3 sin .cos = − θ − × + θ θ 2 2 2 2 1 3cos .sin 3 sin .cos =− θ θ× + θ θ = –3 + 3 = 0 244. The expression (cos 6 θ + sin 6 θ –1) (tan 2 θ + cot 2 θ + 2) + 1 is equal to: (a) 1 (b) –2 (c) 0 (d) –1 SSC CGL (Tier-I) 18/04/2022 (Shift-II) Ans. (b) (cos 6 θ + sin 6 θ-1) (tan 2 θ + cot 2 θ + 2) +1 ( ) { } ( ) 2 2 2 4 4 2 2 cos sin cos sin cos sin 1 tan cot 1 = θ+ θ θ+ − θ θ− θ+ θ +     ( ) 2 2 2 2 2 2 2 2 2 sin cos 1 cos sin 2sin cos cos sin 1 1 sin cos θ+ θ     =× θ+ θ− θ θ− θ θ− +     θ θ   2 2 2 2 1 3sin cos 1 sin cos =− θ θ× + θ θ = –3 + 1 = – 2 245. If cot θ = 3 5 , 0° < θ < 90° then the value of θ θ 2 2 2 2 5 6sec - cosec θ 3 3 4 sec + cosec θ 5 3 , is equal to:

Trigonometry 148 . (a) 1/3 (b) 1 (c) 1/2 (d) 2/3 SSC CHSL –20/10/2020 (Shift-II) Ans : (b) cot θ = 3 5 ⇒ cos 3 sin 5 θ θ = By taking the value of cosθ = 3, sinθ = 5 2 2 2 2 5 6sec θ- cosec θ 3 3 4 sec θ + cosec θ 5 3 = 2 2 2 2 1 5 1 6 3 3 5 3 1 4 1 5 3 3 5     × − ×             × +         = 2 1 3 3 1 4 15 15   −       +     = 1 3 1 3 = 1 246. If tanθ = 2 11 , 0 < θ < 90 0 , then the value of 2 2 2 2 2cosec θ – 3sec θ 3cosec θ + 4sec θ is equal to: (a) 11 45 (b) 11 49 (c) 13 49 (d) 10 49 SSC CHSL –26/10/2020 (Shift-III) Ans. (d) : 2 2 2 2 2cosec θ - 3sec θ 3cosec θ + 4sec θ = ( ) ( ) ( ) ( ) 2 2 2 2 21 cot 31 tan 31 cot 41 tan + θ− + θ + θ+ + θ = 2 2 2 2 2cot 3tan 1 3cot 4 tan 7 θ− − θ+ θ+ = 11 4 2 3 1 4 11 11 4 3 4 7 4 11 × − × − × + × + = 121 24 22 22 363 64 308 44 − − + + = 75 2 10 735 49 × = 247. If A and B are acute angles and SecA = 3; Cot B = 4, then the value of 2 2 2 2 cosec A + sin B cot A + sec B is: (a) 25 261 (b) 322 323 (c) 1 261 (d) 2 SSC CHSL –14/10/2020 (Shift-II) Ans. (b) : sec A = 3, cot B = 4 2 2 2 2 cosec A sin B cot A sec B + + 2 2 2 2 3 1 8 17 1 17 4 8     +         =     +         9 1 8 17 1 17 8 16 + = + 161 8 17 19 16 × = = 322 323 248. If cosA, sinA, cotA are in geometric progression, then the value of tan 6 A – tan 2 A is: (a) 1 (b) 1 3 (c) 1 2 (d) 3 SSC CHSL –13/10/2020 (Shift-II) Ans. (a) : Q cosA, sinA, cotA is in G.P. ∴ 3 2 3 2 1 2 T T sinA cotA sin A cos A T T cosA sinA = ⇒ = ⇒ = (where T = term) 3 2 3 3 3 sin A cos A tan A sec A cos A cos A = ⇒ = ∴ tan 6 A – tan 2 A = sec 2 A – tan 2 A = 1 249. If 11 secA = 3 , then the value of 2 2 2 2 cosec A + tan A sin A + cot A is: (a) 4 9 (b) 11 9 (c) 9 4 (d) 2 11 SSC CHSL–13/10/2020 (Shift-II) Ans. (b) : Given, 11 sec A 3 = 2 11 2 tanA sec A 1 1 9 3 = − = − = sinA = 2 9 2 1 cos A 1 11 11 − = − = ∴ 2 2 2 2 11 2 cosec A tan A 2 9 2 9 sin A cot A 11 2 + + = + + = ( ) ( ) 99 4 18 99 4 22 / / + + = 22 11 18 9 =

Trigonometry 149 . 250. If 7cos 2 θ + 3sin 2 θ = 6, 0° < θ < 90°, then the value of 2 2 2 2 cot 2θ + sec 2θ tan 2θ - sin 2θ is: (a) 49 45 (b) 28 27 (c) 52 27 (d) 26 15 SSC CGL (Tier-I)– 04/03/2020 (Shift-III) Ans. (c) : 7cos 2 θ+ 3sin 2 θ=6 4cos 2 θ + 3(cos 2 θ+sin 2 θ)=6 4cos 2 θ + 3 = 6 4cos 2 θ = 3 cos 2 θ =3/4 θ = 30º Hence 2 0 2 0 2 0 2 0 cot 60 sec 60 tan 60 sin 60 + − 1 4 3 3 3 4 + = − 13 4 52 3 9 27 × = = × 251. If 8sin 2 θ + 2cosθ = 5, 0° < θ < 90°, then the value of tan 2 θ + sec 2 θ – sin 2 θ will be: (a) 431 144 (b) 23 9 (c) 153 72 (d) 305 144 SSC CHSL 12/04/2021 (Shift-I) Ans : (d) 8sin 2 θ + 2cosθ = 5 8–8cos 2 θ + 2cosθ = 5 8cos 2 θ – 6cosθ+4cosθ –3 = 0 2cosθ (4 cosθ – 3) + 1(4 cosθ–3) = 0 (4 cosθ–3) (2cosθ+1) = 0 cosθ = 3 4 cosθ = 1 2 − (invalid) tan 2 θ + sec 2 θ – sin 2 θ 7 16 9 9 16 7 + − 112 256 63 305 144 144 + − = 252. If 3 cos 2 θ - 4sinθ + 1 = 0, 0º < θ < 90º, then tanθ + secθ = ? (a) 2 5 (b) 2 3 (c) 33 (d) 5 SSC CGL (Tier-I) 16/08/2021 (Shift I) Ans. (d) : 3 cos 2 θ – 4sinθ + 1 = 0 3(1–sin 2 θ) – 4sinθ + 1 = 0 3 – 3sin 2 θ – 4sinθ + 1 = 0 3sin 2 θ + 4sinθ – 4 = 0 3sin 2 θ + 6sinθ – 2sinθ – 4 = 0 (sinθ + 2)(3sinθ – 2) = 0 ∴ 3sinθ = 2 2 sin 3 θ= (on taking) sin 1 tan sec cos θ+   θ+ θ=   θ   5/3 4 1 9 = − 5/3 5 3 = 5 = 253. If tanθ = 5, then the value of 2 2 2 2 cosec θ + sec θ cosec θ - sec θ is : (a) 7 5 − (b) 7 5 (c) 3 2 − (d) 3 2 SSC CGL–(Tier-I) 18/08/2021 (Shift II) Ans. (c) : Given, 2 2 2 2 coses θ+sec θ tanθ= 5then the vlue of is: cosec θ-sec θ Q cosec 2 θ = 1 + cot 2 θ sec 2 θ = 1 + tan 2 θ 1 1 cot tan 5 θ= = θ 2 2 2 2 1 cot 1 tan 1 cot 1 tan + θ+ + θ = + θ− − θ 1 2 5 5 1 5 5 + + = − 1 7 5 1 5 5 + = − 36 5 24 5 = − 36 24 =− 3 2 =−

Trigonometry 150 . 254. If cos 2 θ – sin 2 θ – 3cos θ + 2 = 0, 0º < θ < 90º, then what will be the value of sec θ – cos θ θθ θ? (a) 4 3 (b) 1 2 (c) 3 2 (d) 2 3 SSC CHSL 09/08/2021 (Shift-I) Ans. (c) : cos 2 θ – sin 2 θ – 3cos θ + 2 = 0 cos 2 θ –1+ cos 2 θ – 3cos θ + 2 = 0 2 cos 2 θ – 3cos θ + 1 = 0 2 cos 2 θ – 2cos θ – cos θ + 1 = 0 2 cosθ (cos θ –1) –1 (cos θ –1) = 0 (cos θ –1) (2 cos θ –1) = 0 cos θ = 1, 1 2 (Q cos θ = 1, cos θ = 1 2 , sec θ = 1, secθ = 2) by taking value, cos θ = 1 2 ∴ sec θ – cos θ = 1 2 2 − = 3 2 by taking value cos θ = 1 sec θ – cos θ = 1–1 = 0 Hence it is clear that cosθ = 1 2 satisfied answer. Trick:– cos 2 θ – sin 2 θ – 3cosθ + 2 = 0 taking θ = 60º ∴L.H.S = cos 2 60º–sin 2 60º–3cos60º+2 1 3 3 2 4 4 2 = − − + 1 1 0 R.H.S 2 2 =− + = = So, secθ–cosθ=sec60º–cos60º 1 3 2 2 2 = − = 255. 1 + 2tan 2 θ + 2sinθ.sec 2 θ, 0º < θ < 90º, is equal to: (a) 1 cos 1 cos − θ + θ (b) 1 cos 1 cos + θ − θ (c) 1 sin 1 sin − θ + θ (d) 1 sin 1 sin + θ − θ SSC CGL (Tier-I) 16/08/2021 (Shift I) Ans. (d) : 1 + 2tan 2 θ + 2sinθ.sec 2 θ 2 2 2 sin 2sin 1 2. cos cos θ θ + + θ θ 2 2 2 cos 2sin 2sin cos θ+ θ+ θ θ 2 2 2 2 cos sin sin 2sin 1 sin θ+ θ+ θ+ θ − θ 2 2 cos 1 sin θ= − θ     Q ( )( ) 2 1 sin 2sin 1 sin 1 sin + θ+ θ − θ + θ ( ) ( ) ( )( ) 2 2 2 2 2 a b a b 2ab a b a b a b   + = + +    − = − +    Q ( ) ( )( ) 2 1 sin 1 sin 1 sin + θ − θ + θ 1 sin 1 sin + θ ⇒ − θ 256. If 4sin 2 θ = 3(1+cosθ), 0º < θ < 90º, then what is the value of (2tanθ + 4sinθ – secθ)? (a) 3 15 4 − (b) 15 3 4 − (c) 15 3 3 + (d) 4 5 3 − SSC CGL (Tier-I) 11/04/2022 (Shift-I) Ans. (a) 4sin 2 θ = 3(1 + cosθ) 4(1 – cos 2 θ) = 3 (1 + cosθ) 4(1 – cosθ) (1 + cosθ) = 3 (1 + cosθ) 4 – 4 cosθ = 3 cosθ 1 4 = AC 2 = AB 2 + BC 2 4 2 = 1 2 + BC 2 BC = 15 2 tanθ + 4sinθ – secθ = 15 15 4 2 4 1 4 1 × + × − = 3 15 4 − 257. If 117 cos 2 A + 129 sin 2 A = 120 and 170 cos 2 B + 158 sin 2 B = 161, then the value of cosec 2 A sec 2 B is: (a) 16 (b) 9 (c) 1 (d) 4 SSC CHSL –13/10/2020 (Shift-I) Ans. (a) : 117 cos 2 A + 129 sin 2 A = 120 117 cos 2 A + 117 sin 2 A + 12 sin 2 A = 120 117 (cos 2 A + sin 2 A) + 12 sin 2 A = 120 12 sin 2 A = 120 – 117 sin 2 A = 3 12 sin 2 A = 1 4 cosec 2 A = 4---------(i) And 170 cos 2 B + 158 sin 2 B = 161 158 cos 2 B + 12 cos 2 B + 158 sin 2 B = 161 12 cos 2 B + 158 (sin 2 B + cos 2 B) = 161 12 cos 2 B = 161 – 158 cos 2 B = 3 12 cos 2 B = 1 4 ⇒ sec 2 B = 4---------(ii) Multiplying equation (i) and (ii) cosec 2 A sec 2 B = 4×4 = 16

Trigonometry 151 . 258. If 5 cos θ = 13 , then the value of 2 2 tan θ + sec θ is equal to : (a) 233 25 (b) 303 25 (c) 313 25 (d) 323 25 SSC CGL (Tier-II) – 18/11/2020 Ans. (c) cosθ 5 13 sec 13 5 = ⇒ θ= tan 2 θ + sec 2 θ = (sec 2 θ –1) + sec 2 θ = 2sec 2 θ – 1 = 2 2 13 2 1 5 × − 313 25 = 259. If 6 6 0 0 1 sin θ + cos θ = ,0 <θ<90 3 then what is the value of sinθcosθ ? (a) 2 3 (b) 6 6 (c) 2 3 (d) 2 9 SSC CGL–(Tier-I) 18/08/2021 (Shift III) Ans. (c) : 6 6 1 sin cos 3 θ+ θ= ( ) ( ) 3 3 2 2 a b 1 b a ab b + = + − + Q So, (sin 2 θ) 3 + (cos 2 θ) = (sin 2 θ + cos 2 θ) (sin 2 θ – sin 2 θ cos 2 θ + cos 2 θ) 1 1 (1 3 = × ( ) 2 2 1 1 1 sin .cos 3 = × − θ θ 2 2 2 sin cos 3 ∴ θ θ= 2 sin cos 3 θ θ= 260. If 2 2 2 sin θ = 5, tan θ-sin θ then the value of 2 2 2 2 24cos θ -15sec θ 6cosec θ - 7cot θ is: (a) 6 (b) 2 (c) 4 (d) 1 SSC CHSL 06/08/2021 (Shift-III) Ans. (b) : 2 2 2 sin 5 tan sin θ = θ− θ sin 2 θ = 5tan 2 θ – 5sin 2 θ 6sin 2 θ = 5tan 2 θ ( ) 2 2 2 2 5 5 1 cos ,sin 1 sin 1 cos 6 6 6 θ= θ= − = θ= − θ Q 2 2 2 2 24 cos 15sec 6 cosec 7 cot θ− θ θ− θ Q 5 6 24 15 6 5 6 6 7 5 × − × = × − × ( ) 2 2 sin 1 cos θ= − θ Q 20 18 2 1 − = = 261. If sin θ + sin 2 θ = 1, then the value of cos 2 θ + cos 4 θ is equal to : (a) 5 (b) 0 (c) 1 (d) 1 2 SSC CGL (Tier-II) – 18/11/2020 Ans. (c) : Q sin θ + sin 2 θ = 1, cos 2 θ + cos 4 θ = ? sin θ = 1 – sin 2 θ sin θ = cos 2 θ then, cos 2 θ + cos 4 θ =? cos 2 θ + (cos 2 θ) 2 sin θ + sin 2 θ 1 ∴ 2 4 cos cos 1 θ+ θ= 262. If x sin 3 θ + y cos 3 θ = sin θ cos θ and x sin θ = y cos θ, then the value of x 2 + y 2 is: (a) 2 (b) 4 (c) 0 (d) 1 SSC CHSL –20/10/2020 (Shift-I) Ans : (d) Given, x sin θ = y cos θ sin y cos x = θ θ .....(i) ∴ x sin 3 θ + y cos 3 θ = sin θ cos θ xy 3 + yx 3 = xy From equation (i) xy (y 2 +x 2 ) = xy x 2 +y 2 = 1 263. The least value of 8cosec 2 θ + 25sin 2 θ is: (a) 20 2 (b) 30 2 (c) 40 2 (d) 10 2 SSC CHSL –12/10/2020 (Shift-I) Ans. (a) : The minimum value of 8 cosec 2 θ + 25 sin 2 θ 2 a b = × 2 25 8 = × 20 2 = Note : a sin 2 θ + b cosec 2 θ a cos 2 θ + b sec 2 θ a tan 2 θ + b cot 2 θ Minimum value 2 ab = (IV) Problems based on Angular Values of Trigonometric Functions 264. If θ is an acute angle and sinθ = 43 47 , then find the value of cotθ? (a) 43 6 10 (b) 6 10 43 (c) 6 10 47 (d) 47 6 10 SSC CHSL (Tier-I) 14/08/2023 (Shift-IV) Ans. (b): sinθ = 43 47 ∴ cosθ = 2 1 sin − θ

Trigonometry 152 . ⇒ cosθ = 2 43 1 47   −     ⇒ cosθ = 360 47 47 × ⇒ cos θ = 6 10 47 So, cotθ = cos 6 10 / 47 6 10 sin 43 / 47 43 θ = = θ ⇒ cotθ = 6 10 43 265. If cot B = 15 8 , where B is an acute angle, then the value of (secB + tanB) is: (a) 4 5 (b) 3 5 (c) 5 3 (d) 5 4 SSC CHSL (Tier-I) 03/08/2023 (Shift-II) Ans. (c) : cot B = 15 8 tan B = 1 1 cot B 15/8 = ⇒ tan B = 8/15 We know that, sec B = 2 1 tan B + ⇒ sec B = 2 1 (8/15) + ⇒ sec B = 225 64 225 + ⇒ sec B = 17 15 So, sec B + tan B = 17 8 25 15 15 15 + = ⇒ Sec B + tan B = 5 3 266. Find the value of 8sec 2 45° + 20sin 2 30° + 15tan 45°. (a) 43 (b) 28 (c) 32 (d) 36 SSC CHSL (Tier-I) 04/08/2023 (Shift-III) Ans. (d) : 8 sec 2 45º + 20 sin 2 30° + 15 tan 45º = 8 × ( ) 2 2 + 20 ×(1/2) 2 + 15 × (1) = 2 × 8 + 5 + 15 = 36 267. Simplify the following. 7(cot 11º cot 13º cot 19º cot 29º cot 61º cot 71º cot 77º cot 79º) (a) 7 (b) –1 (c) –7 (d) 1 SSC CHSL (Tier-I) 09/08/2023 (Shift-III) Ans. (a) : 0 0 0 0 0 0 0 0 7(cot11 cot13 cot19 cot 29 cot 61 cot 71 cot 77 cot 79 ) 0 0 0 0 0 0 0 0 7 (cot (90 79 ) cot(90 77 ) cot (90 71 ) cot(90 61 ) − − − − 0 0 0 0 cot 61 cot 71 cot 77 cot 79 ) 0 0 0 0 0 0 0 0 7(tan 79 . tan 77 tan 71 tan 61 cot 61 cot 71 cot 77 cot 79 ) ( ) ( ) ( ) 7 tan 79 .cot 79 tan 77 cot 77 tan 71 cot 71 tan 61 cot 61 ° ° ° ° ° ° ° °     = 7 [1 1 1 1] [ tan .cot 1] × ××× θ θ= Q = 7 268. If cos(40 + x) = sin30° then the value of x is? (a) 90º (b) 20º (c) 23º (d) 22º SSC CGL (Tier-I) 25/07/2023 (Shift-IV) Ans. (b) : cos(40º + x) = sin 30º, x = ? cos(40º + x) = sin (90º – 60º) cos(40º + x) = cos 60º [จ sin(90º – θ) = cosθ] 40º + x = 60º x = 20º 269. In a right angle triangle for an acute angle x, If sinx = 3 7 , then find the value of cosx? (a) 3 4 (b) ( ) 2 10 7 (c) 1 3 (d) 2 7 SSC CGL (Tier-I) 26/07/2023 (Shift-II) Ans. (b) : 3 sin x 7 = , cosx = ? we know that cos 2 x = 1 – sin 2 x 9 40 1 49 49 = − = 40 2 10 cos x 49 7 = = 270. What will be the value of tan 25 0 tan 35 0 tan 45 0 tan 55 0 tan 65 0 ? (a) 1 (b) 2 (c) 3 (d) 0 SSC CGL (Tier-I) 21/07/2023 (Shift-II) Ans. (a) : tan 25 0 tan 35 0 tan 45 0 tan 55 0 tan 65 0 = tan (90 0 – 65 0 ) tan (90 0 – 55 0 ) × 1 × tan 55 0 tan 65 0 = cot 65 0 cot 55 0 × 1 × tan 55 0 × tan65º = (tan65º. cot65º) (tan55º. cot55º) × 1 = 1 × 1 × 1 [Q tanθ cotθ = 1] = 1 271. What will be the value of (cos 2 32º – sin 2 58º) ? (a) 0 (b) 1 (c) –1 (d) 1 2 SSC CGL (Tier-I) 19/07/2023 (Shift-IV)

Trigonometry 153 . Ans. (a) : cos 2 32º – sin 2 58º cos 2 (90º – 58º) – sin 2 58º sin 2 58º – sin 2 58º [จ cos (90 0 – θ) = sinθ] = 0 272. If tan 4 θ + θ + θ + θ + tan 2 θ θ θ θ = 1, then the value of 11 (cos 4 θ + cos 2 θ) will be : (a) –11 (b) 0 (c) 8 (d) 11 SSC CGL (Tier-I) 27/07/2023 (Shift-III) Ans. (d) : tan 4 θ + tan 2 θ = 1, then 11 (cos 4 θ + cos 2 θ) = ? tan 2 θ ( tan 2 θ + 1) = 1 tan 2 θ. sec 2 θ = 1 2 2 2 sin 1 1 cos cos θ × = θ θ cos 4 θ = sin 2 θ cos 4 θ = 1 – cos 2 θ [จ sin 2 θ = 1 – cos 2 θ] cos 4 θ + cos 2 θ = 1 11 ( cos 4 θ + cos 2 θ) = 11 × 1 = 11 273. If 3 cosθ = 2 then tan 2 θ.cos 2 θ =? (a) 3 (b) 1 2 (c) 1 4 (d) 1 3 SSC CGL (Tier-I) 14/07/2023 (Shift-I) Ans. (c) : 3 cos 2 θ= 0 cos cos30 θ= 0 30 θ= 2 2 2 0 2 0 tan .cos tan 30 cos 30 θ θ= 2 2 1 3 2 3     = ×         1 3 1 3 4 4 = × = 274. If sin (a + b) = 1 and cos(a – b) = 1 2 , then find the value of a. (a) 30° (b) 45° (c) 75° (d) 15° SSC CGL (Tier-I) 14/07/2023 (Shift-I) Ans. (c): Given that sin (a + b) = 1 sin (a + b) = sin90 0 , a + b = 90 0 ............(i), ( ) 1 cos a b 2 − = cos (a – b) = cos 60 0 a – b = 60 0 ...........(ii) and, from equation (i) & (ii) a + b = 90 0 a – b = 60 0 2a = 150 0 ⇒ a = 75 0 275. Find the value of the given expression. 2 0 2 2 0 2 0 sec (90 - θ) - cot θ 2(sin 35 + sin 55 ) (a) 0.5 (b) 2 (c) 1 (d) 1.5 SSC CGL (Tier-I) 20/07/2023 (Shift-I) Ans. (a) : 2 0 2 2 0 2 0 sec (90 -θ)-cot θ 2(sin 35 +sin 55 ) 2 2 2 0 0 2 0 cosec θ – cot θ 2(sin (90 – 55 )+sin 55 ) = 2 2 [ cos ec θ – cot θ = 1] Q 2 0 2 0 1 = 2(cos 55 + sin 55 ) 2 2 1 = [ cos + sin 1] 2 θ θ= Q = 0.5 276. If is an acute angle and tan(4α – 50°) = cot(50 – α), then find the value of α (in degree). (a) 90º (b) 30º (c) 45º (d) 60º SSC CGL 08/12/2022 (Shift-IV) Ans. (b) : Given that tan (4α–50º) = cot (50º–α) ⇒ tan (4α–50º) = tan [90º–(50º–α)] ⇒ 4α–50º = 40º+α ⇒ 3α = 90º ง α = 30º 277. What is the value of sin28° sin35° sin45° sec62° sec55° ? (a) 1 2 (b) 2 (c) 1/2 (d) 2 SSC CGL 09/12/2022 (Shift-I) Ans. (a) : sin28º .sin35º .sin45º.sec62º.sec55º = sin28º. sec(90º–28º).sin35º.sec(90º–35º).sin45º = sin28º.cosec28º.sin35º.cosec35º. sin45º 1 2 = 278. If tan(α + β) = a, tan (α – β) = b, then the value of tan2α is. (a) a b 1 ab + − (b) a b a ab + + (c) a b 1 ab − − (d) a b 1 ab − + SSC CGL 13/12/2022 (Shift-IV)

Trigonometry 154 . Ans. (a) : Given that tan(α + β) = a, tan(α – β) = b tan2α = tan[(α + β) + (α – β)] = ( ) ( ) ( ) ( ) tan tan 1 tan tan α+β + α−β − α+β α−β = a b 1 ab + − 279. If sec 2θ = cosec (θ – 36°), where θ is an acute angle, then find the value of θ. (a) 46° (b) 20° (c) 32° (d) 42° SSC CGL 05/12/2022 (Shift-I) Ans. (d) : sec2θ = cosec (θ – 36º) sec2θ = sec [90º – (θ – 36º)] 2θ = 90º – θ + 36º 3θ = 126º θ = 126 3 θ = 42º 280. What will be the value of Cos 17 3 π         −                 ? (a) 1 2 (b) 1 (c) 0 (d) 3 2 SSC CGL 06/12/2022 (Shift-II) Ans. (a) : 17π cos 3 −       17 cos 3 π = ( ) cos cos −θ = θ     Q o cos1020 = ( ) cos 3 360 60 = × °− ° o 1 cos 60 2 = = ( ) o cos 360 cos   −θ = θ   281. What will be the value of o o cos 37 sin 53 ? (a) 1 2 (b) 0 (c) 1 (d) 1 2 SSC CGL 06/12/2022 (Shift-II) Ans. (c) : cos37 sin 53 ° ° ( ) o o o cos 37 sin 90 37 = − o o cos 37 cos 37 = =1 282. Which of the following is equal to (sin 30º cos60º– cos30º sin60º) ? (a) cos 30º (b) – sin30º (c) – cos30º (d) sin30º SSC CGL 01/12/2022 (Shift-II) Ans. (b) : sin30º. cos60º–cos30º.sin 60º [ ∵ sin (A –B) = sinA. cosB – cosA. sinB] = sin (30º – 60º) = sin (– 30º) = – sin 30º 283. If sinθ + cosθ = 5 sin(90° – θ), then find the value of cotθ : (a) 5 1 5 − (b) 5 1 3 + (c) 5 1 4 − (d) 5 1 4 + SSC CGL 02/12/2022 (Shift-II) Ans. (d) : sinθ + cosθ = 5 sin (90 – θ) sinθ + cosθ = 5 cos θ On dividing both sides by cosθ, sin cos 5 cos cos cos cos θ θ θ + = θ θ θ tan θ + 1 = 5 tan θ = 5 –1 cot θ = 1 5 1 5 1 5 1 5 1 5 1 + + × = − − + = 5 1 4 + 284. ( ) ( ) 1 1 + 1+cos 90º-θ 1 - cos 90º -θ = ________. (a) 1 (b) 2sec 2 θ (c) 0 (d) 2tan 2 θ SSC CGL 03/12/2022 (Shift-I) Ans. (b) : ( ) ( ) 1 1 1 cos 90º 1 cos 90º + + −θ − −θ = 1 1 1 sin 1 sin + + θ − θ = ( )( ) 1 sin 1 sin 1 sin 1 sin − θ+ + θ + θ − θ = 2 2 1 sin − θ 2 2 cos = θ = 2sec 2 θ 285. If sec3θ = cosec(4θ – 15°), then find the value of tan3θ. (a) 3 (b) –1 (c) 1 (d) 1 3 SSC CGL 06/12/2022 (Shift-IV) Ans. (c): ( ) sec3 cosec 4 15 ο θ= θ− ( ) cosec 90 - θ = secθ     Q ( ) ( ) 0 o cosec 90 3 cosec 4 15 −θ= θ− ⇒ 0 o 90 3 4 15 − θ= θ− ⇒ o 7 105 θ= ∴ o 15 θ= Now, tan 3θ · o tan 3 15 × · tan 45 o · 1

Trigonometry 155 . 286. What is the value of cos 2 15°? (a) ( ) 2 3 2 + (b) ( ) 2 3 4 + (c) ( ) 1 3 2 + (d) ( ) 2 3 + SSC CGL 06/12/2022 (Shift-IV) Ans. (b) : Formula → cos2A = 2cos 2 A – 1 On taking A = 15° 2 cos 2 15 2cos 15 1 × °= °− 2 cos 30 1 2cos 15 °+ = ° 2 3 1 2cos 15 2 + = ° 2 2 3 cos 15 4 + °= 287. Find the value of the following. sin 25° sin 65° – cos 25° cos 65°. (a) 4 (b) 1 (c) 0 (d) 40 SSC CGL (Mains) 02/03/2023 Ans. (c) : sin25º.sin65º–cos25º.cos65º = sin25º.sin65º–cos(90º–65º).cos(90º–25º) = sin25º.sin65º– sin65º.sin25º = sin25º.sin65º–sin25º.sin65º = 0 288. What is the value of 0 0 0 0 3sin58 3sin42 + cos32 cos48 ? (a) 7 (b) 9 (c) 6 (d) 8 SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (c) : 0 0 0 0 3sin58 3sin 42 cos32 cos 48 + ( ) ( ) ( ) { } 0 0 0 0 3sin 90 32 3sin 90 48 sin 90 cos cos 32 cos 48 − − = + −θ = θ Q 0 0 0 0 3cos32 3cos 48 cos 32 cos 48 = + = 3 + 3= 6 289. Find the value of 0 0 0 0 0 0 0 0 2 0 2 0 sin23 cos67 + sec52 sin38 + cos23 sin67 + cosec52 cos38 cosec 20 - tan 70 (a) 4 (b) 3 (c) 0 (d) 2 SSC CGL 11/04/2022 (Shift-I) Ans. (b) : 0 0 0 0 0 0 0 0 2 0 2 0 sin23 cos67 + sec52 sin38 + cos23 sin67 + cosec52 cos38 cosec 20 - tan 70 ( ) ( ) ( ) ( ) ( ) 2 2 sin 90° - 67° .cos67° + sec52°.sin 90° - 52° + cos 90° - 67° .sin67° + cosec52°.cos 90° -52° cosec 90 20 tan 70 °− °− ° 2 2 cos67 .cos67 sec52 .cos52 sin67 .sin67 cosec52 .sin52 sec 70 tan 70 ° °+ ° °+ ° °+ ° ° °− ° 2 0 2 0 2 2 sec .cos 1 cos 67 1 sin 67 1 cos ec .sin 1 1 sec tan 1 θ θ=   + + +   = θ θ=     θ− θ=   = sin 2 67 0 + cos 2 67 0 + 1 + 1 = 1 + 1 + 1 {sin 2 θ + cos 2 θ = 1} =3 290. What is the value of 0 0 0 0 0 0 cos50 3cosec80 + - 2cos50 .cosec40 sin40 sec10 ? (a) 2 (b) 4 (c) 5 (d) 3 SSC CGL (Tier-II) 08/08/2022 (Shift-I) Ans. (a) : 0 0 0 0 0 0 cos50 3cos ec80 2cos50 .cosec40 sin 40 sec10 + − =cos50 0 .cosec40 0 +3cosec80 0 .cos10 0 – 2cos50 0 . cosec40 0 if, α+ β = 90 0 then, cosα . cosecβ = 1 = 1 + 3 – 2 = 2 291. Find the value of the following expression. 3 0 3 0 2 0 2 0 0 tan 45 + 4cos 60 2cosec 45 - 3sec 30 + sin30 (a) 1 2 + (b) 4 3 (c) 3 4 (d) 3 SSC CGL 13/04/2022 (Shift-I) Ans. (d) : 3 0 3 0 2 0 2 0 0 tan 45 4cos 60 2cosec 45 3sec 30 sin 30 + − + () ( ) 3 3 2 2 1 1 4 2 2 1 2 2 3 2 3   + ×     =   × − × +     1 1 4 8 4 1 2 2 3 3 2 + × = × − × + 1 1 2 1 4 4 2 + = − + 3/2 1/2 = = 3 292. If A = 10°, then find the value of ( ) ( ) 0 0 12sin3A + 5cos 5A - 5 9A 9sin - 4cos 5A + 10 2 .

Trigonometry 156 . (a) ( ) 6 2 5 9 2 2 + + (b) ( ) 6 2 5 9 2 2 − − (c) ( ) 9 2 2 6 2 5 − + (d) ( ) 6 2 5 9 2 2 + − SSC CGL 18/04/2022 (Shift-III) Ans. (d) : Given that, A = 10 0 ( ) ( ) 0 0 12sin3A 5cos 5A 5 9A 9sin 4cos 5A 10 2 + − = − + ( ) ( ) 0 0 0 0 0 0 12sin3 10 5cos 5 10 5 9 10 9sin 4cos 5 10 10 2 × + × − = × − × + 0 0 0 0 12sin30 5cos45 9sin 45 4cos60 + = − 1 1 12 5 2 2 1 1 9 4 2 2 × + × = × − × 6 2 5 2 9 2 2 2 + = − 6 2 5 9 2 2 + = − 293. 2 0 2 0 2 0 2 0 2 0 0 0 0 0 0 0 4tan 30 +sin 30 cos 45 +sec 48 -cot 42 cos37 sin53 + sin37 cos53 + tan18 tan72 What will be the value of (a) 35 24 (b) 35 48 (c) 59 48 (d) 49 24 SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (c) : 2 2 2 2 2 4 tan 30 sin 30 cos 45 sec 48 cot 42 cos 37 sin 53 sin 37 cos 53 tan18 tan 72 °+ ° °+ °− ° ° °+ ° °+ ° ° ( ) 2 2 2 2 2 1 1 1 4 . sec 48 tan 48 2 3 2 sin53 .sin53 cos53 .cos53 tan18 cot18       × + + °− °             = ° °+ ° °+ °× ° ∴ {sec 2 θ – tan 2 θ = 1, sin 2 θ + cos 2 θ =1, tanθ.cotθ=1} ( ) 2 0 2 0 1 1 1 4 1 3 4 2 sin 53 cos 53 1 × + × + = + + 4 1 1 3 8 2   + +     = 59 48 = 294. ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 0 2 0 2 0 2 0 0 2 0 0 3 cot 47 -sec 43 - 2 tan 23 -cosec 67 cosec 68 +θ -tan θ+61 - tan 22 -θ +cot 29 -θ What will be the value of (a) 0 (b) 5 (c) 1 (d) –1 SSC CGL (Tier-II) 29/01/2022 (Shift-I) Ans. (d) : 2 2 2 2 2 2 3(cot 47º – sec 43º ) – 2(tan 23º – cosec 67º ) cosec (68º + ) – tan( + 61º ) – tan (22º – ) + cot(29º – ) θ θ θ θ ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 0 2 0 2 0 2 0 2 0 0 3 cot 47 sec 43 2 tan 23 cos ec 67 cos ec 68 tan 61 tan 22 cot 29 − − − +θ − θ+ − −θ + −θ ∴ tan (90 – θ) = cotθ sec 2 θ – tan 2 θ = 1 tan 2 θ – sec 2 θ = –1 a + b = 90 0 , tanA = cotB cosec 2 θ – cot 2 θ = 1 ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 0 2 0 2 0 2 2 0 3 tan 43 sec 43 2 tan 23 sec 23 cos ec 68 tan 61 tan 22 cot 29 − − − = +θ − θ+ − −θ + −θ ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 tan 43 sec 43 2 tan 23 sec 23 cosec 68 tan 61 cot 68 tan 61 °− °− °− ° = °+θ − θ+ − +θ + θ+ ° ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 tan 43°-sec 43° -2 tan 23°-sec 23° = cosec 68° + θ - cot 68° + θ ( ) ( ) ( ) ( ) 2 2 2 2 2 2 -3 -tan 43 sec 43 2 tan 23 sec 23 = cos ec 68 cot 68 °+ °+ − °+ ° °+θ − °+θ 3 2 1 −+ = = –1 295. If sinθ = 3 2 and θ θ θ θ is an acute angle, then find the value of cos3θ. θ. θ. θ. (a) 1 (b) 0 (c) ½ (d) –1 SSC CHSL –10/06/2022 (Shift-I) Ans. (d) : Given that, 3 sin 2 θ= sin θ = sin 60º θ = 60º cos 3θ = cos 3 × 60º = cos 180º = cos (180 + 0º) = –cos 0º = –1 296. Using the formula x 1 - cosx tan = 2 sinx , find the value of tan22.5°. (a) 2 1 − (b) 3 2 2 + (c) 3 2 2 − (d) 2 1 + SSC CHSL –10/06/2022 (Shift-I)

Trigonometry 157 . Ans. (a) : x 1-cosx tan = 2 sinx On putting x = 45º 45 1 cos 45 tan 2 sin 45 ° − ° = ° 1 1 2 tan 22.5º 1 2 − = 2 1 2 1 2 − = tan 22.5º 2 1 = − 297. What will be the value of (1 + tan 10º)( 1 + tan 35º). (a) 1 (b) 2 (c) 1/2 (d) 3/4 SSC CHSL –25/05/2022 (Shift-III) Ans. (b) : According to the question, (1 + tan 10º)( 1 + tan 35º) ⇒ tan (10º + 35º) = tan 10º + tan 35º 1 – tan 10º tan 35º 1 – tan 10º. tan 35º = tan 10° + tan 35º [Q tan 45º = 1] 1 = tan 10º + tan 35º + tan 10º . tan 35º 1 + 1 = tan 10º + tan 35º + tan 10º . tan 35º + 1 2 = 1 + tan 10º + tan 35º (1 + tan 10º) 2 = (1 + tan 10º)(1 + tan 35º) 298. If cos2A = sin75°, what will be the smallest positive value of A? (a) 37.5º (b) 15º (c) 7.5º (d) 30º SSC CHSL –27/05/2022 (Shift-II) Ans. (c) : Given that cos 2A = sin 75º sin (90º – 2A) = sin 75º 90º – 2A = 75º 2A = 90º – 75º A = 7.5º 299. cot75° = 2 – 3 , then find the value of cot 15°. (a) 2 + 3 (b) 3 + 1 (c) 2 – 3 (d) 3 – 1 SSC CHSL –24/05/2022 (Shift-III) Ans. (a) : cot 75º = 2 – 3 , cot 15º = ? cot 15º = cot (60º – 45º) cot 60º.cot 45º 1 cot 45º cot 60º + = − 1 1 3 1 1 3 3 3 1 3 1 3 1 1 3 3 + + + = = = − − − 1 3 3 1 3 1 3 3 3 1 3 1 3 1 + + + + + = × = − − + 2 3 4 2( 3 2) 3 2 2 2 + + = = ⇒ + 300. What will be the value of 2 0 2 0 2 0 2 0 5cos 62 + 5cos 28 - 21 7sin 35 + 7sin 55 +1 ? (a) –2 (b) 2 (c) –3 (d) 3 SSC CGL 20/04/2022 (Shift-III) Ans. (a) : 2 0 2 0 2 0 2 0 5cos 62 5cos 28 -21 7sin 35 7sin 55 1 + + + Q sin (90–θ) =cosθ, cos (90–θ) = sinθ ( ) ( ) 2 0 2 0 2 0 2 0 5 cos 90 28 5cos 28 21 7 sin 90 55 7sin 55 1   − + −   =   − + +   2 0 2 0 2 0 2 0 5sin 28 5cos 28 21 7cos 55 7sin 55 1 + − = + + { } 2 2 sin cos 1 θ+ θ= Q 5 21 16 2 7 1 8 − − = = =− + 301. What will be the value of ( )( ) 0 0 0 0 2 0 2 0 2 0 2 0 cos9 + sin81 sec9 + cosec81 cosec 71 + cos 15 - tan 19 + cos 75 (a) 1 (b) –3 (c) 4 (d) 2 SSC CPO SI 9/11/2022 (Shift I st ) Ans. (d): ( ) ( ) 0 0 0 0 2 0 2 0 2 0 2 0 cos9 sin 81 sec9 cos ec81 cosec 71 cos 15 tan 19 cos 75 + + + − + ∴ sin(90–θ) = cosθ cosec (90–θ) = secθ sin 2 θ + cos 2 θ = 1 cosec 2 θ – cot 2 θ = 1 2 2 2 2 [cos9º sin(90º –9º )][sec9º cosec(90º –9º )] cosec 71º + cos (90º –75º ) – tan (90º –71º ) + cos 75º + + = ( )( ) 0 0 0 0 2 0 2 0 2 0 2 0 cos9 cos 9 sec9 sec9 cos ec 71 sin 75 cot 71 cos 75 + + = + − + 0 0 2cos9 2 2 cos 9 × = × = 2 302. Find the value of sin 4 30º + cos 4 30º – sin 25º cos 65º – sin 65º cos 25º. (a) 5 8 (b) 3 8 − (c) 0 (d) 13 8 SSC CPO SI 9/11/2022 (Shift I st ) Ans. (b) : sin 4 30 0 +cos 4 30 0 –sin25 0 cos65 0 –sin65 0 cos25 0 ( ) ( ) 4 4 0 0 0 0 0 0 1 3 sin 90 65 cos65 sin 65 cos 90 65 2 2     = + − − − −           ( ) 2 0 2 0 1 9 cos 65 sin 65 16 16 = + − + 10 1 16 = − 5 3 1 8 8 − = − =

Trigonometry 158 . 303. If secθ and sinθ (0° < θ < 90°) are the roots of the equation 2 6x – kx + 6=0 then the value of k is: (a) 33 (b) 32 (c) 2 3 (d) 3 SSC CHSL –14/10/2020 (Shift-I) Ans. (a) : 2 6x kx 6 0 − + = Multiple of roots = secθ.sinθ = 6 6 tanθ = 1, θ = 45 0 Sum of roots · secθ + sinθ = k 6 sec45 0 + sin45 0 = k 6 1 k 2 2 6 + = 3 k 2 6 = k = 33 304. If tan 2 x – 3sec 2 x + 3 = 0, then the value of x(0 ≤ x ≤ 90 0 ), is: (a) 30 0 (b) 0 0 (c) 45 0 (d) 60 0 Ans. (b) : 2 2 tan x 3sec x 3 0 − + = 2 2 tan x sec x 1   = −   Q 2 2 sec x 1 3sec x 3 0 − − + = – 2sec 2 x = –2 sec 2 x = 1 sec x = sec0° x = 0 0 305. Which of the following is irrational number ? (1) tan30 0 tan60 0 (2) sin30 0 (3) tan45 0 (4) cos30 0 (a) (3) (b) (1) (c) (2) (d) (4) SSC CHSL (Tier-I) 10/07/2019 (Shift-II) Ans. (d) (i) tan 30 0 tan 60 0 1 3 1 3 = × = (ii) sin 30 0 1 2 = (iii) tan 45 0 =1 (iv) cos 30 0 = 3 2 Hence, it is clear that 0 3 cos30 2 = which is represented a irrational number. 306. Find the value of o o o o tan 60 – tan 15 1 + tan 60 tan 15 (a) 1 (b) 1 2 (c) 3 2 (d) 1 2 SSC CHSL –12/10/2020 (Shift-II) Ans. (a) : o o o o tan 60 – tan15 1+ tan 60 tan15 Compare from formula. tan A – tan B tan (A B) 1 + tan A tan B = − = tan (60 0 – 15 0 ) = tan 45 0 = 1 307. If cosA cosA cosecA + 1 cosecA 1 + − = 2, 0º ≤ A ≤ 90º, then A is equal to: (a) 45º (b) 90º (c) 30º (d) 60º SSC CHSL –21/10/2020 (Shift-I) Ans. (a) From question, cosA cosA + 2 cosecA +1 cosecA -1 = 2 cosA(cosecA -1) + cosA(cosecA +1) 2 cosec A 1 = − 2 cotA - cosA + cotA + cosA 2 cot A = 2 2cotA 2 cot A = tan A = 1 ⇒ tan A = tan45º ∴ A = 45º 308. What is the value of [2 sin (45+θ) sin (45-θ)]/cos 2θ? (a) 0 (b) tan 2θ (c) cot 2θ (d) 1 SSC CGL (Tier-II) 19-02-2018 Ans. (d) : ( ) ( ) 0 0 2sin 45 .sin 45 cos 2 +θ −θ θ [ ] 2sin A.sin B cos(A B) cos(A B) = − − + Q [ ] cos(45 45 ) cos(45 45 ) cos 2 +θ− +θ − +θ+ −θ = θ 0 cos 2 cos90 cos 2   θ− =   θ   cos 2 cos 2 θ = θ = 1 309. What is the value of [sin (90–A) + cos (180– 2A)]/[cos (90–2A) + sin (180–A]? (a) sin (A/2) cosA (b) cot (A/2) (c) tan (A/2) (d) sin A cos (A/2) SSC CGL (Tier-II) 21-02-2018 Ans. (c) : ( ) ( ) ( ) ( ) 0 0 0 0 sin 90 A cos 180 2A cos 90 2A sin 180 A   − + −     − + −   ( ) ( ) cos A cos2A sin2A sin A − = +

Trigonometry 159 . A 2A 2A A 2sin .sin 2 2 2A A 2A A 2sin .cos 2 2 + −             = + −             3A A 2sin .sin 2 2 3A A 2sin .cos 2 2             =             A sin 2 A cos 2 = A tan 2 = 310. What is the value of [sin (90°–10θ) – cos (π– 6θ)]/[cos(π/2–10θ) – sin (π–6θ)] ? (a) tan 2θ (b) cot 2θ (c) cot θ (d) cot 3θ SSC CGL (Tier-II) 20-02-2018 Ans. (b) : ( ) ( ) ( ) ( ) sin 90 10 cos 6 cos 90 10 sin 6  − θ − π− θ       − θ − π− θ      cos10 cos6 sin10 sin6 θ+ θ = θ− θ 10 6 10 6 2cos .cos 2 2 10 6 10 6 2cos .sin 2 2 θ+ θ θ− θ = θ+ θ θ− θ cos 2 sin 2 θ = θ = cot2θ 311. What is the value of (cos 40 0 – cos 140 0 )/(sin 80 0 + sin 20 0 ) ? (a) 2 3 (b) 2/ 3 (c) 1/ 3 (d) 3 SSC CGL (Tier-II) 21-02-2018 Ans. (b) : ( ) ( ) 0 0 0 0 cos40 cos140 sin80 sin 20 − + 0 0 0 0 0 0 0 0 40 140 140 40 2sin .sin 2 2 80 20 80 20 2sin .cos 2 2     + −         =     + −         0 0 0 0 2sin 90 .sin 50 2sin50 .cos30 = 0 0 sin 90 1 cos30 3 2 = =       2 3 = 312. What is the value of [1–tan (90–θ) + sec (90 – θ)]/[tan (90–θ) + sec (90 –θ) + 1] ? (a) cot (θ/2) (b) tan (θ/2) (c) sin θ (d) cos θ SSC CGL (Tier-II) 21-02-2018 Ans. (b) : ( ) ( ) ( ) ( ) 0 0 0 0 1 tan 90 sec 90 tan 90 sec 90 1   − −θ + −θ     −θ + −θ +   [ ] [ ] 1 cot cosec cot cosec 1 − θ+ θ = θ+ θ+ cos 1 1 sin sin cos 1 1 sin sin θ   − +   θ θ   = θ   + +   θ θ   ( ) ( ) sin cos 1 cos sin 1 θ− θ+ = θ+ θ+ 2 2 2sin / 2.cos /2 1 2sin /2 1 2sin / 2.cos /2 2 cos /2 1 1 θ θ − + θ + = θ θ + θ − + ( ) ( ) 2sin /2 cos /2 sin /2 2cos /2 sin /2 cos /2 θ θ + θ = θ θ + θ sin /2 cos /2 θ = θ tan /2 = θ 313. What is the value of {sin (90–x) cos [π–(x–y)]} + {cos (90–x) sin [π–(y–x)]} ? (a) –cos y (b) –sin y (c) cos x (d) tan y SSC CGL (Tier-II) 20-02-2018 Ans. (a): {sin(90–x) cos [π – (x–y)]} + {cos(90–x) sin[π–(y–x)]} = – cosx cos (x–y) + sinx sin (y–x) = –cosx cos (x–y) – sinx sin (x–y) = –[cosx cos(x–y) + sinx sin (x–y)] = –cos[x–(x–y)] = –cos [x–x+y] = – cos y 314. What is the value of - 2 2 2 tan 60° 2sin 45° cos24°cos37°cosec53°cos60°cosec66° + sin 60° (a) 1 4 5 (b) 1 3 5 (c) 2 (d) 1 SSC CHSL 15/04/2021 (Shift-II) Ans : (b) 2 2 2 tan 60º 2sin 45º cos 24º cos 37º cos ec53º cos 60º cos ec66º sin 60º − + ( ) 2 2 1 3 2 2 cos 24º cos37º 1 3 sin 66º sin 53º 2 4   − ×     = × × + 3 1 2 1 3 cos 24 cos 37 1 3 2 4 cos 24 cos 37 2 4 − = =     + × × +         8 3 1 5 5 = = 315. What is the value of (32 cos 6 x – 48 cos 4 x + 18 cos 2 x–1)/ [4 sin x cosx sin (60 –x) cos (60–x) sin (60 + x) cos (60 + x) ? (a) 4 tan 6x (b) 4 cot 6x (c) 8 cot 6x (d) 8 tan 6x SSC CGL (Tier-II) 21-02-2018

Trigonometry 160 . Ans. (c) : ( ) ( ) ( ) ( ) ( ) 6 4 2 0 0 0 0 32cos x 48cos x 18cos x 1 4 sin x cos x.sin 60 x .cos 60 x sin 60 x cos 60 x − + − = − − + + 0 0 1 sin x.sin(60 x)sin(60 x) sin 3x 4 − + = Q and cosx.cos(60°–x). cos(60°+x) = 1 cos3x 4 ∴ 6 4 2 2(16cos x 24cos x 9cos x 1) 1 1 4 sin 3x cos 3x 4 4 − + − × × = 3 2 2 3 2 (4cos x) (3cos x) 2 4cos x 3cosx 1 1 2sin 3x.cos3x 8   + − × × −   × = 3 2 2(4cos x 3cosx) 1 1 sin 6x 8 − − ] = 2 8 2cos 3x 1 sin 6x × − = 8 cos6x 8cot6x sin 6x × = 316. What is the value of ( ) ( ) ( ) ( )                       3 3 4cos 90 - A sin 90 + A - 4sin 90 + A cos 90 - A 180 + 8A cos 2 ? (a) 1 (b) –1 (c) 0 (d) 2 SSC CGL (Tier-II) 17-2-2018 Ans. (b) : ( ) ( ) ( ) ( ) { } 0 3 0 0 3 0 0 4cos 90 A sin 90 A 4sin 90 A cos 90 A 180 8A cos 2     − + − + −       +     ( ) 3 3 0 4sin A.cos A 4cosA.sin A cos 90 4A − = + {∴cos(90 0 –θ) = sinθ cos (90 + θ) = –sinθ sin (90–θ) = cosθ sin (90+θ) = cosθ} ( ) 2 2 4sinAcosA cos A sin A sin 4A − = − 2.2sinAcosA.cos2A sin4A = − 2.sin 2A.cos 2A sin 4A = − {Qsin2A = 2sinA cosA} sin 4A 1 sin4A = =− − 317. What is the value of 2 sin 15 0 cos 15 0 – 4 sin 3 15 0 cos 15 0 ? (a) 3/ 2 (b) 3/2 (c) 3/4 (d) 1/2 SSC CGL (Tier-II) 9-3-2018 Ans. (c) : 2sin15 0 .cos15 0 – 4 sin 3 15 0 .cos15 0 = 2sin15 0 .cos15 0 [1–2sin 2 15 0 ] = sin30 0 × cos30 0 , [2sinθ.cosθ = sin2θ, 1–2sin 2 θ = cos2θ] 1 3 3 2 2 4 = × = 318. What is the value of cos 15 0 + cos 105 0 ? (a) 3 (b) 1/ 2 (c) 3/2 (d) 1/ 3 SSC CGL (Tier-II) 9-3-2018 Ans. (b) : cos15 0 + cos105 0 0 0 120 90 2 cos . cos 2 2 = [cosC+cosD = 2cos C D C D . cos 2 2 + − ] 1 1 1 2 2 2 2 = × × = 319. What is the value of cos 15 0 – cos 165 0 ? (a) 3/ 2 (b) ( ) 2/ 3 1 − (c) ( ) 3 1/ 2 + (d) ( ) 3 1 /2 + SSC CGL (Tier-II) 9-3-2018 Ans. (c) : C D D C cos C cos D 2sin .sin 2 2 + − − = 0 0 0 0 180 150 cos15 cos165 2sin .sin 2 2 − = = 2 × 1 × sin 75 0 sin 75 0 = sin(30 0 + 45 0 ) = sin30 0 cos45 0 + cos30 0 . sin45 0 1 3 3 1 2 2 2 2 2 2 + = + = 0 0 3 1 3 1 cos15 cos165 2 2 2 2   + + ∴ − = × =       320. If P + Q + R = 60 0 , then what is the value of cosQ cos R (cos P – sin P) + sin Q sinR (sinP – cos P) ? (a) 1/2 (b) 3/2 (c) 1/ 2 (d) 2 SSC CGL (Tier-II) 9-3-2018

Trigonometry 161 . Ans. (a) : Given : P + Q + R = 60º ------(i) Because P, Q and R are variables. ∴ From equation (i) we can take P = Q = 0º and R = 60º cosQ. cosR (cosP – sinP) + sinQ.sinR (sinP–cosP) ( ) 1 1 1 0 0 2 = × − + 1 2 = 321. What is the value of [1–tan (90–θ)] 2 /[cos 2 (90– θ)] – 1 ? (a) – sin2θ (b) – cos2θ (c) cos 2θ (d) sin 2θ SSC CGL (Tier-II) 9-3-2018 Ans. (a) : ( ) ( ) 2 0 2 0 1 tan 90 cos 90   − −θ   −θ –1 By value putting, θ = 45 0 ( ) 0 2 0 1 tan 45 1 cos 45 − = − = 0–1 = –1 Form the option (a), –sin2θ = – sin90 0 = –1 Hence, option (a) is correct. 322. What is the value of [(sin 59 0 cos 31 0 + cos 59 0 sin31 0 ) ÷ (cos 20 0 cos 25 0 – sin 20 0 sin 25 0 )] ? (a) 1/ 2 (b) 2 2 (c) 3 (d) 2 SSC CGL (Tier-II) 18-02-2018 Ans. (d) : 0 0 0 0 0 0 0 0 sin59 .cos31 cos 59 sin 31 cos20 .cos25 sin 20 sin 25 + − sin(A+B) = sinA.cosB + cosA. sinB cos(A+B) = cosA.cosB – sinA.sinB ( ) ( ) 0 0 sin 59 31 sin 90 1 2 1 cos 20 25 cos 45 2 + ⇒ = = = + 323. What is the value of cos (90–B) sin (C–A) + sin (90 + A) cos (B + C) – sin (90–C) cos (A+B) ? (a) 1 (b) sin (A + B – C) (c) cos (B + C – A) (d) 0 SSC CGL (Tier-II) 18-02-2018 Ans. (d): cos(90–B) sin(C–A) + sin(90+A)cos (B+C) – sin( 90–C) cos(A+B) = sinB sin (C–A) + cosA cos(B+C) – cosC cos(A+B) = sinB[sinC.cosA – cos C.sinA] + cosA [cosB.cosC– sinB.sinC] – cos C [cosA.cosB – sinA.sinB] = 0 324. 2cosθ + secθ – 2 2 = 0, where θ is an acute angle. Find the value of θ. (a) 60° (b) 30° (c) 15° (d) 45° SSC CHSL 12/04/2021 (Shift-III) Ans : (d) 2cos sec 2 2 0 θ+ θ− = let, θ=45º L.H.S 2 cos 45º sec 45º 2 2 = × + − 2 2 2 2 = + − 2 2 2 2 0 R.H.S = − = = θ=45º 325. The value of cos8ºcos24º cos60º cos66º cos82º sin82º sin66º sin60º sin8º sin24º is (a) 1 (b) 1 2 (c) 1 3 (d) 0 SSC CHSL 09/08/2021 (Shift-III) Ans. (c) : cos 8º cos 24º cos 60º cos 66º cos 82º sin 82º sin 66º sin 60º sin 8º sin 24º cos 8º cos 24º cos 60º cos 66º cos82º cos 8º cos 24º sin 60º cos 66º cos82º = × × × × ( ) ( ) sin 90 cos −θ = θ Q 1 cos 60º 1 2 sin 60º 3 3 2 ∴= = = 326. In a right-angled triangle ABC right angled at C, sin A = sin B. What is the value of cos A? (a) 3 2 (b) 1 (c) 1 2 (d) 1 2 SSC CHSL 09/082021 (Shift-II) Ans. (c) : sinA=sinB Let ∠A =∠B=45º sin45º=sin45º 1 1 2 2 = 1 cos A cos 45º 2 ∴ = = 327. If sin 2 x = 3 cos 2 x and 0° ≥ x ≤ 90°, then what is the value of x 2 ? (a) 30° (b) 45° (c) 15° (d) 22.5° SSC CHSL 12/08/2021 (Shift-II) Ans. (a) : sin 2 x=3cos 2 x tan 2 x = 3 tan x 3 tan 60º = = x = 60º x 60 30º 2 2 ∴ = =

Trigonometry 162 . 328. What is the value of sin 2 60 + tan 2 45 + sec 2 45 – cosec 2 30? (a) – 1 4 (b) 1 4 (c) 4 (d) –4 SSC CHSL 12/04/2021 (Shift-II) Ans : (a) sin 2 60º+tan 2 45+ sec 2 45-cosec 2 30 () ( ) ( ) 2 2 2 2 3 1 2 2 2   = + + −       3 1 2 4 4 = + + − = 3 1 1 4 4 − = − = 329. If 3 3 sec θ + 4 tanθ 3 tan θ + 3 sec θ = 2, 0° < θ < 90°, then the value of cosθ will be: (a) 3 2 (b) 1 4 (c) 1 2 (d) 1 2 SSC CHSL 13/04/2021 (Shift-III) Ans.(d) : 3 3sec 4 tan 2 3tan + 3 sec θ+ θ = θ θ 3 sec = 2tan θ θ 3 sin sin 60º 2 θ= = 60º θ= 1 cos cos 60º 2 ∴ θ= = 330. What number should be subtracted from 4(sin 4 60°+cos 4 30°)–(tan 2 45°–cot 2 30°)+cos 2 45° – cosec 2 45° + sec 2 60° to get 2? (a) 5 (b) 4 (c) 7 (d) 3 SSC CHSL 13/04/2021 (Shift-III) Ans.(c) : ( ) ( ) 4 4 2 2 2 2 2 4 sin 60º cos 30º tan 45º cot 30º cos 45º cosec 45º sec 60º ? 2 + − − + − + − = ( ) ( ) 4 4 2 2 2 2 2 ? 4 sin 60º cos 30º tan 45º cot 30º cos 45º cos 45º sec 60º 2 = + − − + − + − () ( ) ( ) () 4 4 2 2 2 2 3 3 1 ? 4 1 3 2 2 2 2 2 2             = + − − + − + −                         [ ] 9 9 1 ? 4 1 3 2 4 2 16 16 2   = + − − + − + −     18 1 ? 4 2 16 2   = + +     9 5 ? 2 2 = + 14 ? 2 = ? = 7 331. If 2 2cos θ = 3(1 - sinθ), 0º < θ < 90º , then what is the value of (tan2θ + cosec3θ + sec2θ)? (a) √3 + 1 (b) 3 1 − (c) 1 3 3 + (d) 1 3 3 − SSC CHSL 04/08/2021 (Shift-III) Ans. (b) : 2 cos 2 θ = 3 (1–sinθ) (Q 0 0 < θ < 90 0 ) Putting, θ = 30 0 3 1 2 31 4 2   × = −     3 3 2 2 = tan 2 θ + cosec3θ – sec2θ = tan60 0 + cosec 90 0 – sec60 0 3 1 2 = + − 3 1 = − 332. For A = 30º, Find the value of: 2 2 3A -3sin 2A + 2sec A - tan 2 1 sin3A 3 (a) 7 4 − (b) 7 36 − (c) 57 4 (d) 11 4 SSC CHSL 05/08/2021 (Shift-III) Ans. (a) : Given, ∠A = 30º 2 2 3 3A 3sin 2A 2sec tan 2 1 sin A 3 − + − 2 2 3sin 60º 2sec 30º tan 45º 1 sin 90º 3 − + − = × 3 4 9 8 7 3 2 1 1 4 3 4 3 12 1 1 1 3 3 3 − − −× + × − + − = = = 7 4 − = 333. Find the value of θ, if 2 sec θ + (1 - 3 )tanθ - (1 + 3)=0 , where θ is an acute angle. (a) 60º (b) 30º (c) 45º (d) 15º SSC CHSL 05/08/2021 (Shift-III) Ans. (a) : ( ) ( ) 2 sec 1 3 tan 1 3 0 θ+ − θ− + = From option (a), Let θ = 60º ( ) ( ) 2 sec 60º 1 3 tan 60º 1 3 0 + − − + =

Trigonometry 163 . ( ) 4 1 3 3 1 3 0 + + − − = 4 3 4 3 0 + − − = 0=0 ⇒L.H.S = R.H.S. 334. If 3cosθ = 2sin 2 θ, 0° < θ < 90°, then what is the value of (tan 2 θ+sec 2 θ–cosec 2 θ) ? (a) 7 3 − (b) 17 3 − (c) 17 3 (d) 7 3 SSC CHSL 10/08/2021 (Shift-III) Ans. (c) : 3cosθ = 2sin 2 θ Letθ = 60º, 3 L.H.S 3 cos 60º 2 = × = 2 3 3 R.H.S 2 sin 60º 2 4 2 = × = × = ∴ L.H.S=R.H.S Now, tan 2 θ + sec 2 θ–cosec 2 θ = tan 2 60º+sec 2 60º–cosec 2 60º 4 3 4 3   = + −     17 3 = 335. The value of cosec (58°+θ)–sec(32°–θ) + sin15° sin35° sec55° sin30° sec75° is: (a) 0 (b) 1 (c) 1/2 (d) 2 SSC CHSL 12/08/2021 (Shift-I) Ans. (c) : cosec (58°+θ)–sec(32°–θ) + sin15° sin35° sec55° sin30° sec75° cos(90 ) sin sin(90 ) cos −θ = θ     −θ = θ   Q =cosec (58°+θ)–sec(32°–θ) + sin15 sin 35 sin 30 sin15 sin 35 ° ° × × ° ° ° = cosec (58°+θ)–sec(32°–θ) + 1 2 = 1 1 1 sin(58 ) cos(32 ) 2 − + °+θ °−θ = 1 1 1 sin(58 ) sin(58 ) 2 − + °+θ °−θ = 1 2 336. Solve the following equation and find the value of θ. 0 3cotθ + tanθ - 2 3 = 0, 0 < θ < 90 (a) 30 0 (b) 45 0 (c) 15 0 (d) 60 0 SSC CHSL 04/08/2021 (Shift-II) Ans. (d) : 3cot tan 2 3 0 θ+ θ− = 3cot tan 2 3 θ+ θ= 3 tan 2 3 tan θ+ = θ put the value of θ = 60° in L.H.S. 0 0 3 tan 60 tan 60 + 3 3 3 3 2 3 3 + = + = = R.H.S ∴ L.H.S = R.H.S Hence, the required value of θ is 60° 337. The value of - cos60°cos30° sin60°sin30° is: (a) 1 (b) 1 2 (c) 0 (d) 3 2 SSC CHSL 10/082021 (Shift-II) Ans. (c) : cos 60º cos 30º sin 60º sin 30º − ( ) ( ) ( ) cos 60 30º cos A B cos A cos B sin A sin B = + + = − Q cos90º 0 = = 338. If 2 2 sec α + 4cos α = 4 and 0º ≤ α ≤ 90º, then find the value of α. (a) 30º (b) 0º (c) 45º (d) 60º SSC CGL (Tier-I) 21/04/2022 (Shift-III) Ans : (c) sec 2 α + 4cos 2 α = 4 0º ≤ α ≤ 90º 2 2 1 4 cos 4 cos + α= α 4cos 4 α – 4cos 2 α + 1 = 0 (2cos 2 α-1) 2 = 0 2cos 2 α = 1 2 1 cos 2 α= 1 cos 2 α= 45º α= 339. If (2cosA +1) (2cosA – 1) = 0, 0º < A ≤ 90º, then find the value of A. (a) 90º (b) 45º (c) 30º (d) 60º SSC CGL (Tier-I) 21/04/2022 (Shift-I) Ans : (d) (2cosA +1) (2cosA–1) = 0 0º< A ≤90º 4cos 2 A–1 = 0 2 1 cos A 4 = 1 cos A 2 = A = 60º

Trigonometry 164 . 340. If A = 30º, What is the value of :     2 8sinA + 11cosecA - cot A 10cos2A ? (a) 1 5 5 (b) 3 4 5 (c) 2 4 5 (d) 4 3 5 SSC CGL (Tier-I) 21/04/2022 (Shift-I) Ans : (b) Given, A= 30º 2 8sin A 11cosecA cot A ? 10 cos 2A + −     = 1 8 11 2 3 2 1 10 2 × + × − = × 4 22 3 5 + − = 23 4 5 5 3 = = 341. If cot 2 α + tan 2 α = 2, 0º ≤ α ≤ 90º, then find the value of a. (a) 0º (b) 45º (c) 60º (d) 90º SSC CGL (Tier-I) 19/04/2022 (Shift-III) Ans. (b) Given, cot 2 α + tan 2 α = 2 Put θ = 45º 1 + 1 = 2 2 = 2 L.H.S = R.H.S Hence, θ = 45º 342. If A = 60º, what is the value of: A 10sin + 8cosA 2 ? 3A 7sin - 12cosA 2 (a) 10 (b) 12 (c) 9 (d) 7 SSC CGL (Tier-I) 19/04/2022 (Shift-I) Ans. (c) Given, A = 60º The given expression is– A 10sin 8cosA 2 3A 7sin 12 cos A 2 + ⇒ − 60 10sin 8cos60º 2 60 7sin3 12 cos 60º 2 + ⇒ × − 10sin 30º 8cos60º 7 sin 90 12cos60º + ⇒ − 1 1 10 8 2 2 1 7 1 12 2 × + × ⇒ ×− × ⇒9 343. If A = 10º, what is the value of: ( ) 12 sin 3A + 5cos (5A - 5º ) 9A 9sin - 4cos 5A + 10º 2 (a) ( ) 6 2 5 9 2 2 + − (b) ( ) 6 2 5 9 2 2 − − (c) ( ) ( ) 9 2 2 6 2 5 − + (d) ( ) 6 2 5 9 2 2 + + SSC CGL (Tier-I) 18/04/2022 (Shift-III) Ans. (a) Given, that– A = 10º 12sin 3A 5cos(5A 5º ) ? 9A 9sin 4 cos(5A 10º ) 2 + − ∴ = − + ( ) ( ) 12sin 30º 5cos 50º 5º 90º 9sin 4cos 50º 10º 2 + − ⇒ − + 1 12 5 cos 45º 2 9sin 45º 4 cos 60º × + ⇒ − 1 6 5 2 1 1 9 4 2 2 + × ⇒ × − × ( ) 6 2 5 9 2 2 + ⇒ − 344. Find the value of following expression : 3 3 2 2 tan 45° + 4cos 60° 2cosec 45° - 3 sec 30° + sin 30° (a) 3/4 (b) 1+ 2 (c) 4/3 (d) 3 SSC CGL (Tier-I) 13/04/2022 (Shift-I) Ans. (d) 3 3 2 2 tan 45° + 4cos 60° 2 cosec 45° - 3sec 30° + sin 30° 1 1 4 8 4 1 2 2 3 3 2 + × = × − × + 1 1 2 1 4 4 2 + = − +

Trigonometry 165 . 3 2 1 2 = = 3 345. If 2k sin30º cos30º cot60º 2 2 cot 30º sec60º tan45º = cosec 45ºcosec30º , then find the value of k. (a) 3 2 (b) 3 (c) 1 (d) 6 SSC CGL (Tier-I) 12/04/2022 (Shift-III) Ans.(b) From question, 2k sin 30º cos 30º cot 60º = 2 2 cot 30º sec 60º tan 45º cosec 45º cosec 30º 2k × ( ) ( ) 2 2 3 2 1 1 3 1 2 2 3 2 2 × × × × = × k 3 2 2 = ∴ k = 3 Hence, option (b) is correct. 346. tan 2 A + 5secA = 13, where 0 < A < 90º. Solve for A (in degrees). (a) 0 (b) 60 (c) 45 (d) 30 SSC CGL (Tier-I) 12/04/2022 (Shift-III) Ans.(b) From question, tan 2 A + 5secA = 13, 0º < A < 90º ⇒ 2 2 sin A 5 13 cos A cos A + = sin A 1 tan A ,secA cos A cos A   = =     Q 2 2 sin A 5cosA 13 cos A + = sin 2 A + 5cosA = 13cos 2 A (1– cos 2 A) + 5cosA – 13cos 2 A = 0 { } 2 2 sin A cos A 1 + = Q –14 cos 2 A +5 cosA +1 = 0 14 cos 2 A –5 cosA –1 = 0 14 cos 2 A – (7 – 2) cosA –1 = 0 14 cos 2 A –7 cosA +2 cosA –1 = 0 7 cosA (2 cosA –1) +1 (2 cosA –1) = 0 (2 cosA –1) (7 cosA +1) = 0 ∴ 2 cosA –1 = 0 ⇒ 2 cosA = 1 ⇒ cosA = 1 2 ⇒ cosA = cos 60º A = 60º Short Method : By option (b), A = 60º tan 2 60º + 5sec 60º = 13 ⇒ ( ) 2 3 5 2 13 + × = ⇒ 3 + 10 = 13, 13 = 13 Hence, option (b) is correct. 347. If cos (A – B) 3 = 2 and secA = 2, 0º ≤ A ≤ 90º, ≤ B ≤ 90º, then what is the measure of B? (a) 60º (b) 0º (c) 30º (d) 90º SSC CGL (Tier-I) 11/04/2022 (Shift-II) Ans. (c) Given secA = 2 A = 60º cos (A – B) 3 2 = A – B = 30º 60º – B = 30º B = 30º 348. What is the value of : 2 8 3sin30º tan60º –3cos0º +sin 45º +2cosº 30? (a) 15 (b) 12 (c) 9 (d) 18 SSC CGL (Tier-I) 11/04/2022 (Shift-II) Ans. (b) 2 2 8 3 sin 30º tan60º 3cos0º 3sin 45º 2cos 30º − + + 2 2 1 1 3 8 3 3 3 1 3 2 2 2 2     × × − ×+ × + ×           = 12 – 3+ 3 2 + 3 2 = 9 + 3 = 12 349. If A lies between 45° and 540°, and sinA = 0.5, what is the value of A/3 in degrees? (a) 170° (b) 175° (c) 160° (d) 165° SSC CHSL 15/04/2021 (Shift-I) Ans. (a) : sin A = 0.5 = 1 2 = 30° Where (45° < A < 540°) A = 30°, 150°, 390°, 510 0 A 3 = 0 0 510 170 3 = 350. If tan θ + 3 cotθ – 2 3 = 0, 0 0 < θ < 90 0 then what is the value of (cosec 2 θ + cos 2 θ) ? (a) 2 3 (b) 19 12 (c) 14 3 (d) 11 12 SSC CGL–(Tier-I) 24/08/2021 (Shift I)

Trigonometry 166 . Ans. (b) : tan θ + 3 cot θ – 2 3 0 = By putting value, θ = 60 0 0 0 tan 60 3cot 60 2 3 + − 3 3 2 3 0 = + − = ∴cosec 2 θ + cos 2 θ = cosec 2 60 0 + cos 2 60 0 2 2 2 1 2 3     = +         4 1 19 3 4 12 = + = 351. Find the value of 0 0 0 2 0 2 0 2 0 8sin30 sin260 - 4sin90 - sec 45 tan 45 -cot 30 (a) 1 2 − (b) 5 2 (c) 3 4 (d) 3 2 SSC CGL–(Tier-I) 13/08/2021 (Shift II) Ans. (d) : Find value of 0 2 0 0 2 0 2 0 2 0 8sin 30 sin 60 4sin90 sec 45 tan 45 cot 30 − − − 1 3 8 4 2 3 6 3 2 4 1 3 2 2 = × × − − − − = = − − − 3 2 = 352. The value of 2 0 2 0 2 0 2 0 2 0 0 0 0 0 tan 30 +sin 90 +cot 60 +sin 30 cos 45 sin60 cos30 - cos60 sin30 is: (a) 47 12 (b) 37 12 (c) 43 12 (d) 25 12 SSC CGL–(Tier-I) 16/08/2021 (Shift III) Ans. (c) : 2 0 2 0 2 0 2 0 2 0 0 0 0 0 tan 30 +sin 90 +cot 60 +sin 30 cos 45 sin60 cos30 - cos60 sin30 ( ) 1 1 1 1 43 43 1 43 3 3 4 2 24 24 1 sin 60º 30º sin 30º 12 2   + + + ×     = = = = − 353. The value of 4 (sin 4 30 0 + cos 4 30 0 ) – 3(sin 2 45 0 – 2cos 2 45 0 ) is : (a) 0 (b) 4 (c) 2 (d) 1 SSC CGL–(Tier-I) 16/08/2021 (Shift III) Ans. (b) 4 4 2 2 4(sin 30º cos 30º ) 3(sin 45º 2cos 45º) + − − 1 9 1 1 4 3 2 16 16 2 2     = + − × − ×         10 3 4 2 = + 10 6 16 4 4 4 + = = = 354. The value of ( ) 0 0 0 0 0 2 0 2 0 0 tan13 tan36 tan45 tan54 tan77 2sec 60 sin 60 - 3cos60 + 2 is : (a) 1 10 (b) 1 4 − (c) 1 4 (d) 1 10 − SSC CGL–(Tier-I) 20/08/2021 (Shift III) Ans. (a) ( ) 0 0 0 0 0 2 0 2 0 0 tan13 tan36 tan45 tan54 tan77 2sec 60 sin 60 -3cos60 + 2 is: 1 3 1 2 4 3 2 4 2 =   × − × +     1 3 6 8 8 4 = − +       1 5 8 4 = × 1 10 = 355. Find the value of cot 25 0 cot 35 0 cot 45 0 cot 55 0 cot 65 0 . (a) 1 3 (b) 3 (c) 3 2 (d) 1 SSC CGL–(Tier-I) 13/08/2021 (Shift I) Ans. (d) : cot 25 0 cot 35 0 cot 45 0 cot 55 0 cot 65º = ? = cot 25º. cot 65º cot 45 0 cot 35 0 . cot 55 0 (cot 45 0 =1) Q cot A. cot B = 1 = cot25 0 .cot65 0 cot45 0 cot35 0 .cot55 0 = 1 × 1 × 1 = 1 356. For θ : 0º < θ < 90º 3 sec θ + 4 cos θ = 4√3, find the value of (1 – sin θ + cos θ θθ θ). (a) 1 2 3 2 + (b) 1 3 2 + (c) 1 3 2 − (d) 1 2 3 2 − SSC CHSL 10/08/2021 (Shift-I) Ans. (b) : 3secθ + 4cos θ = 4 3 By value putting, θ = 30° L.H.S= 2 3 3 4 2 3 × + × = 2 3 2 3 4 3 + = =L.H.S=R.H.S Hence 1–sinθ + cosθ = 1–sin30°+cos30° = 1– 1 3 1 3 2 2 2 + + =

Trigonometry 167 . 357. Evaluate the following expression. 2 2 2 2 2 2 2 2 2 3(cot 46º -sec 44º ) 2cos 60º tan 33º tan 57º + 2(sin 28º +sin 62º ) sec (90º -θ) - cot θ (a) –1 (b) 1 (c) –2 (d) 2 SSC CHSL 10/08/2021 (Shift-I) Ans. (a) : 2 2 2 2 2 2 2 2 2 3(cot 46º sec 44º ) 2cos 60º tan 33º tan 57º + 2(sin 28º +sin 62º ) sec (90º θ) cot θ - -- = 2 2 2 2 2 2 2 2 2 1 2× × tan 33º cot 33º 3(cot 46º cosec 46º ) 2 + 2(sin 28º +cos 28º ) cosec θ cot θ       - - [Q sec(90°–θ) = cosecθ, sin(90°–θ) = cosθ and tan (90°–θ) = cotθ] = ( ) 1 1 3 1 2 2 1 1 × ×− + × [Q cot 2 θ–cosec 2 θ = –1] = 3 1 2 2 − + = –1 358. If 2 0 0 2cos 3sin , 0 90 , θ= θ <θ< θ= θ <θ< θ= θ <θ< θ= θ <θ< then the value of 2 2 1 cosce cot 2         θ− θ θ− θ θ− θ θ− θ                 is: (a) –1 (b) 1 4 (c) 1 2 (d) 0 SSC Sel. Post Phase VIII (H.L.) 09.11.20 (Shift-I) Ans. (a): 2 2cos 3sin θ= θ Taking θ = 30 0 , 2 3 1 3 1 2 3 2 3 2 2 4 2     × = × ⇒ × = ×           3 3 L.H.S R.H.S 2 2 ∴ = ⇒ = ∴ Putting the value of θ = 30 0 in 2 2 1 cosec cot 2   θ− θ     2 0 2 0 1 cosec 30 cot 30 2 = × − ( ) 2 2 1 (2) 3 2 = × − = 2 – 3 = –1 359. If cos x = 3 2 − and 3 x , 2 π π< < π< < π< < π< < then find the value of 4cot 2 x – 3cosec 2 x : (a) 0 (b) 8 (c) 1 (d) 2 SSC CHSL 09/07/2019 (Shift-II) Ans. (a): cos x = 3 3 , x 2 2 − π π< < cos x = cos 210º x = 210º 4 cot 2 x – 3 cosec 2 x = 4 × cot 2 210º – 3 cosec 2 210º ( ) ( ) 2 0 0 2 0 0 4.cot 180 30 3cosec 180 30 = + − + = 4 × 3 – 3 × 4 = 0 360. If ( ) 2 0 2 0 0 2 0 2 2 cosec 39 tan 51 sin 90 tan 56 y 3 − − − − − − − − − − − − 2 0 y tan 34 3 = the value of y is : (a) 2 3 − (b) 1 (c) –1 (d) 2 3 SSC CHSL 08/07/2019 (Shift-I) Ans. (b) : ( ) 2 0 2 0 0 2 0 2 0 2 y 2 cosec 39 tan 51 sin 90 tan 56 ytan 34 3 3 − − − = ( ) ( ) 2 0 2 0 0 0 2 0 2 0 0 2 2 cosec 39 tan 90 39 sin 90 tan 56 3 y y tan 90 56 3   ⇒ − − − −   − = 2 0 2 0 0 2 0 2 0 2 2 cosec 39 cot 39 sin 90 tan 56 y 3 y .cot 56 3   ⇒ − − −   = [] 2 y 21 1 y 3 3 ⇒ − ×− = 2 2 cosec cot 1 θ− θ= 2 y 2 y 3 3 ⇒ − = + 4 4y 3 3 ⇒ = y 1 ⇒ = 361. If 4(cosec 2 66 0 – tan 2 24 0 ) + 1 2 sin90 0 – 4tan 2 66 0 y tan 2 24 0 y 2 = then the value of y is : (a) 1 2 − (b) 1 2 (c) –1 (d) 1 SSC CHSL 08/07/2019 (Shift-II) Ans. (d) : 4(cosec 2 66 0 – tan 2 24 0 ) + 1 2 sin90 0 – 4tan 2 66 0 y tan 2 24 0 y 2 = 2 0 2 0 2 0 2 0 2 0 1 sin 24 1 y 4 1 4 cot 24 y tan 24 2 2 sin 66 cos 24   − + ×− × × =     2 0 2 0 2 0 2 0 2 0 1 sin 24 1 1 y 4 4 y tan 24 2 2 cos 24 cos 24 tan 24   − + − × × =    

Trigonometry 168 . 2 0 2 0 1 sin 24 1 y 4 4y 2 2 cos 24   − + − =     2 0 2 0 cos 24 1 y 4 4y 2 2 cos 24   + − =     1 y 4 4y 2 2 + − = y 1 = 362. If 4(cosec 2 65 0 – tan 2 25 0 ) – sin 90 0 – tan 2 63 0 .y.tan 2 0 y 27 2 = , then value of Y is : (a) –1 (b) 2 (c) 1 2 − (d) 1 SSC CHSL (Tier-I)-08/07/2019 (Shift-III) Ans. (b) : ( ) 2 0 2 0 0 2 0 2 0 y 4 cosec 65 tan 25 sin90 tan 63 y.tan 27 2 − − − = 2 0 2 0 0 0 2 0 0 2 0 y 4 Cosec 65 tan (90 65 ) Sin90 tan (90 27 ).y.tan 27 2   ⇒ − − − − − =   ( ) 2 0 2 0 0 2 0 2 0 y 4 cosec 65 cot 65 sin90 cot 27 .y.tan 27 2 ⇒ − − − = y 4 1–1–y 2 ⇒ × = y 3–y 2 ⇒ = ⇒ 6 – 2y = y ⇒ 6 = 3y ⇒ y = 6/3 y = 2 363. If θ is Acute Angle, and given that sec 2 θ + 4tan 2 θ = 6 then find the value of θ ? (a) 60 0 (b) 45 0 (c) 0 0 (d) 30 0 SSC CHSL 10/07/2019 (Shift-II) Ans. (b) : 2 2 sec 4 tan 6 θ+ θ= 2 2 1 tan 4 tan 6 ∴ + θ+ θ= ( ) 2 2 sec 1 tan θ= + θ 2 5 tan 5 θ= () ( ) 2 2 2 0 tan 1 tan 45 θ= = 0 45 θ= 364. If θ = 9 0 then find the value of cotθ cot2θ cot3θ cot4θ cot5θ cot6θ cot7θ cot8θ cot9θ is : (a) 3 (b) 1 (c) 3 1 − (d) 1 3 SSC CHSL 10/07/2019 (Shift-III) Ans. (b) : θ = 9 0 ( ) 0 cot 90 tan 1 tan cot   −θ = θ     θ=   θ   Q = cotθ cot2θ cot3θ cot4θ cot5θ cot6θ cot7θ cot8θ cot9θ = cot9 0 cot18 0 cot27 0 cot36 0 cot45 0 cot54 0 cot63 0 cot72 0 cot81 0 = (cot9 0 . cot81 0 ) × (cot18 0 . cot72 0 ) × (cot27 0 . cot63 0 ) × (cot36 0 . cot54 0 ) × cot45 0 = cot (90 0 – 81 0 ) cot81 0 × cot72 0 . cot (90 0 – 72 0 ) × cot (90 0 – 63 0 ) cot63 0 × cot(90 0 – 54 0 ) ×cot 54 0 × cot45 0 0 0 0 0 0 0 0 0 tan 81 .cot 81 cot 72 .tan 72 cot 63 .tan 63 cot 54 .tan 54 1 = × × × × = 1 365. If θ is a positive acute angle and tan2θ tan3θ = 1, then the value of θ is: (a) 36° (b) 60° (c) 45° (d) 18° SSC CHSL –20/10/2020 (Shift-III) Ans : (d) tan2θ . tan3θ = 1 tan2θ = 1 tan 3θ tan2θ = cot3θ [∴tan(90 – θ) = cot θ] 2θ + 3θ = 90° 5θ = 90° θ = 18° 366. 0 0 0 θ 90 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ kesâ efueS θ keäÙee nw, peyeefkeâ 3cosθ + sinθ = 1 (a) 45 0 (b) 30 0 (c) 90 0 (d) 0 0 SSC CHSL (Tier-I) 11/07/2019 (Shift-I) Ans. (c) : 3 cos sin 1 θ+ θ= ( ) () 2 2 3 1 2 + = By dividing by 2 in both sides, 3 1 1 cos sin 2 2 2 θ+ θ= 0 0 1 cos .cos30 sin .sin30 2 θ + θ = [cos(A–B) = cosA.cosB + sinA.sinB] cos (θ – 30 0 ) = cos60 0 θ – 30 0 = 60 0 θ = 90 0 367. 2 0 2 0 0 0 2 0 2 0 2 0 sin 31 sin 59 tan 29 cot 61 cosec 61 sec 35 cot 55         + + − + − + − + −         −         of simplify value is : (a) –1 (b) 1 2 (c) 0 (d) 1 SSC CHSL 11/07/2019 (Shift-II) Ans. (c): 2 0 2 0 0 0 2 0 2 0 2 0 sin 31 sin 59 tan 29 cot 61 61 sec 35 cot 55 cosec   + + −   −   ( ) ( ) ( ) 2 0 0 2 0 0 0 0 2 0 2 0 0 2 0 sin 90 59 sin 59 tan 90 61 cot 61 1 tan 35 cot 90 35 cosec 61 − + = + − + − − −

Trigonometry 169 . ( ) 2 0 2 0 2 0 2 0 2 0 2 0 cos 59 sin 59 cot 61 1 cot 61 1 tan 35 tan 35 + = + − + + − 2 0 2 0 1 cot 61 1 cot 61 1 = + − − = 1 – 1 = 0 368. (sin 2 36º + tan 2 60º + sec 2 30º + sin 2 54º) is equal to : (a) 14 3 (b) 17 3 (c) 5 (d) 16 3 SSC CHSL 11/07/2019 (Shift-III) Ans. (d) : sin 2 36º + tan 2 60º + sec 2 30º + sin 2 54º = sin 2 (90º – 54º) + ( ) 2 2 2 2 3 sin 54º 3   + +     2 2 4 cos 54º sin 54º 3 3 = + + + 4 1 3 3 = + + 16 3 = 369. If 2sin 2 θ + 3sinθ –2 = 0, (0º < θ < 90º) then find the value of θ is ? (a) 90º (b) 45º (c) 30º (d) 60º SSC CHSL 11/07/2019 (Shift-III) Ans. (c) : 2sin 2 θ + 3sinθ – 2 = 0 (0º < θ < 90º) ( ) 2 3 3 4 2 2 sin 2 2 −± − × − θ= × 3 25 sin 4 −± θ= 3 5 sin 4 −± θ= 1 sin 2 θ= sinθ = sin30º θ = 30º 370. The value of x, if 2sin 2 x = 2 – 3sinx, is: (a) 4 π (b) 6 π (c) 2 π (d) 3 π SSC CHSL –16/10/2020 (Shift-III) Ans. (b) : 2 sin 2 x = 2–3 sin x 2 sin 2 x + 3 sin x – 2 = 0 2 sin 2 x + 4 sin x – sin x – 2 = 0 2 sinx (sin x + 2) –1 (sin x + 2) = 0 (sin x + 2) (2 sin x – 1) = 0 sin x + 2 = 0, 2 sin x – 1 = 0 sin x ≠ –2, sin x = 1 2 = sin 30 o ⇒ x = 6 π 371. If 2sin 2 θ + 5cosθ – 4 = 0, 0 0 < θ < 90 0 then find value of cotθ + cosecθ is : (a) 2 3 (b) 33 2 (c) 3 2 (d) 3 SSC CHSL 05/07/2019 (Shift-I) Ans. (d) : 2 0 0 2sin θ 5cosθ 4 0, 0 θ 90 + − = < < 2 2(1 cos θ) 5cosθ 4 0 − + − = 2 2 2cos θ 5cosθ 4 0 − + − = 2 2cos θ 5cosθ 2 0 − + = By multiplying from–1 2 ( 5) ( 5) 4 2 2 cos θ 2 2 −− ± − − × × = × 5 3 cos θ 4 ± = cos 2 θ= By taking the positive sigh. ∴ 1 cos θ 2 = By taking the sigh. 0 cos θ cos 60 = 0 θ 60 = then, 0 0 cot θ cosec θ cot 60 cosec 60 + = + 1 2 3 3 = + 3 3 = 3 = 372. If tan x = cot (65 0 + 9x) then the value of x is : (a) 2.0 0 (b) 2.5 0 (c) 1.5 0 (d) 1.0 0 SSC CHSL 04/07/2019 (Shift-II) Ans. (b) : ( ) 0 tan x cot 65 9x = + ( ) 0 0 tan x tan 90 65 9x   ⇒ = − +   x 25 9x ⇒ = − 10x 25 ⇒ = 0 25 x 2.5 10 = = 373. If 1 cos x 2 − = and 3π π x , 2 < < < < < < < < then the value of 2tan 2 x + 3cosec 2 x is : (a) 8 (b) 16 (c) 4 (d) 10 SSC CHSL 05/07/2019 (Shift-III) Ans. (d) : 1 cos x 2 =− ( ) 0 0 cos x cos 180 60 = −

Trigonometry 170 . 0 cos x cos120 = 0 x 120 = 2 2 2tan x 3cosec x + 2 0 2 0 2 tan 120 3cosec 120 = + ( ) ( ) 2 2 0 0 0 0 2 tan 180 60 3 cosec 180 60     = − + −     ( ) 2 0 2 0 2 tan 60 3cosec 60 = − + 4 2 3 3 3 = × + × = 10 374. If 1 cos x 2 − = and 3 x 2 π π< < π< < π< < π< < then the value of 4tan 2 x + 3cosec 2 x is : (a) 10 (b) 4 (c) 8 (d) 16 SSC CHSL /07/2019 (Shift-II) Ans. (d) : 1 cos x 2 − = Q and 3 x 2 π π< < cos x cos 3 π   = π+     4 cos x cos 3 π = 4 x 3 π = 2 2 4 tan x 3cosec x ∴ + 2 2 4 4 4 tan 3cosec 3 3 π π = + ( ) 2 2 2 4 3 3 3 −   = +     4 4 3 3 3 = × + × 12 4 = + = 16 375. If ( ) 2 0 2 0 0 2 0 2 6 sec 59 cot 31 sin 90 3tan 56 y 3 − + − − + − − + − − + − 2 0 y tan 34 3 = then the value of y is : (a) –2 (b) 2 (c) 2 3 (d) 2 3 − SSC CHSL /07/2019 (Shift-II) Ans. (b) : ( ) 2 0 2 0 0 2 0 2 0 2 6 sec 59 cot 31 sin 90 3tan 56 3 y y tan 34 3 − + − = ( ) ( ) 2 0 2 0 2 0 2 0 0 2 61 tan 59 cot 31 1 3y tan 56 tan 3 y 90 56 3 + − + ×− × − = ( ) 0 tan 90 cot   −θ = θ   ( ) 2 0 0 2 0 2 0 2 0 2 61 tan 90 31 cot 31 3y tan 56 3 y cot 56 3   + − − + −   × = () 2 y 61 3y 3 3 + − = 2 y 6 3y 3 3 + = + 20 10y 3 3 = 10y = 20 y = 2 376. If 1 1 4secθ, 1 sinθ 1 sinθ + = + = + = + = − + − + − + − + 0 0 < θ < 90 0 then the value of (3cotθ + cosecθ) is : (a) 4 3 (b) 5 3 (c) 5 3 3 (d) 2 3 3 SSC CHSL 03/07/2019 (Shift-II) Ans. (c) : 1 1 4secθ 1 sin θ 1 sin θ + = − + ( )( ) 1 sin θ 1 sin θ 4secθ 1 sin θ 1 sin θ + + − ⇒ = − + 2 2 4secθ 1 sin θ ⇒ = − 2 2 4secθ cos θ ⇒ = 2 2sec θ 4secθ ⇒ = secθ 2 ⇒ = 0 secθ sec 60 ⇒ = 0 θ 60 ⇒ = 3cotθ cosec θ ∴ + 0 0 3cot 60 cosec 60 ⇒ + 1 2 3 3 3 ⇒ × + 3 2 5 53 3 3 3 + ⇒ = = 377. The value of ( ) 2 0 2 0 0 2 0 2 6 sec 59 cot 31 sin 90 3tan 56 3 − − − − − − − − − − − − 2 0 y y tan 34 3 = is : (a) 2 3 (b) 2 3 − (c) 8 5 − (d) 8 5 SSC CHSL 05/07/2019 (Shift-III)

Trigonometry 171 . Ans. (d) : ( ) 2 0 2 0 0 2 0 2 2 y 6 sec 59 cot 31 sin 90 3tan 56 ytan 34 3 3 − − − = ( ) ( ) 2 0 2 0 0 0 2 0 2 0 0 2 6 sec 59 cot 90 59 sin 90 3tan 56 3 y y tan 90 56 3   ⇒ − − − −   − = 2 0 2 0 0 2 0 2 0 2 6 sec 59 tan 59 sin 90 3tan 56 3 y .y cot 56 3   ⇒ − − −   = 2 2 sec θ tan θ 1, 2 y 6 1 3y 3 3 tan θ.cot θ 1   − = ⇒ − ×− =   =   Q 16 10y 3 3 ⇒ = 16 8 y 10 5 ⇒ = = 378. The value of ( ) ( ) 2 2 o o 2 2 cosec 70 - tan 20 tan 63 -θ -cot 27 +θ + sec 37 - cot 53 is: (a) 0 (b) 2 (c) 3 (d) 1 SSC CHSL –19/10/2020 (Shift-I) Ans. (d) : ( ) ( ) 2 o 2 o o o 2 o 2 o cosec 70 tan 20 tan 63 cot 27 sec 37 cot 53 − −θ − +θ + − = ( ) ( ) 2 o 2 o o o 2 o 2 o cosec 70 cot 70 tan 63 tan 63 sec 37 tan 37 − −θ − −θ + − where ( ) ( ) 2 2 2 2 cot 90 tan tan 90 cot , sec tan 1 cos ec cot 1 −θ = θ       −θ = θ θ− θ=     θ− θ=     Q = 0 + 1 1 1 = 379. The value of ( ( ( ) ) )( ( ( ) ) ) 0 0 0 0 0 0 0 0 cos 9 sin 81 sec 9 cosec 81 sin 56 sec 34 cos 25 cosec 65 + + + + + + + + + : (a) 2 (b) 1 (c) 1 2 (d) 4 SSC CHSL 02/07/2019 (Shift-III) Ans. (a) : ( )( ) 0 0 0 0 0 0 0 0 cos9 +sin81 sec9 + cosec81 sin 56 .sec 34 + cos 25 .cosec 65 ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 0 0 0 0 0 0 cos 90 81 + sin 81 sec 90 81 + cosec 81 sin 56 .sec 90 56 + cos 90 65 .cosec 65 − − = − − ( )( ) 0 0 0 0 0 0 0 0 sin 81 + sin 81 cosec81 + cosec81 sin 56 .cosec56 + sin 65 .cosec 65 = ( )( ) 0 0 0 0 0 0 2sin81 2 cosec 81 1 1 sin 56 × +sin65 × sin 56 sin 65 = 0 0 2 2sin81 sin 81 1 1 × = + 4 2 = = 2 380. The value of ( ) 0 0 0 0 0 2 0 2 0 0 tan13 tan37 tan45 tan53 tan77 2cosec 60 cos 60 - 3cos60 + 2 is: (a) 2 (b) 1 (c) 3 2 (d) 1 2 SSC CHSL 01/07/2019 (Shift-III) Ans. (d) : ( ) 0 0 0 0 0 2 0 2 0 0 tan13 tan 37 tan 45 tan 53 tan 77 2cosec 60 cos 60 3cos60 2 − + ( ) ( ) 0 0 0 0 0 0 0 2 2 tan 90 77 .tan 90 53 .tan45 .tan53 .tan77 2 1 1 2. 3 2 2 2 3 − − =       − × +               0 0 0 0 0 cot 77 .cot 53 .tan 45 .tan 53 .tan77 4 1 3 2 2 3 4 2 =   × × − +     0 0 0 0 0 1 1 tan 45 tan 53 tan 77 tan 77 tan 53 8 1 6 8 3 4 × × × × = − +       1 8 3 3 4 = × 1 2 = 381. If sin (θ + 30º) = 3 12 , then the value of θ is equal to: (a) 45º (b) 30º (c) 15º (d) 60º SSC CHSL –21/10/2020 (Shift-II) Ans. (b) Given, sin (θ + 30º) = 3 12 sin (θ + 30º) = 3 3 2 2 3 × sin (θ + 30º) = sin60º θ + 30º = 60º θ = 30º

Trigonometry 172 . 382. If 3(cot 2 θ – cos 2 θ) = 1 – sin 2 θ, 0º < θ < 90º, then θ is equal to: (a) 60 0 (b) 45 0 (c) 30 0 (d) 15 0 SSC CPO-SI 13/12/2019 (Shift-I) Ans. (a) 3(cot 2 θ – cos 2 θ) = 1–sin 2 θ 2 2 2 cos 3 cos sin   θ − θ   θ   = cos 2 θ 3cos 2 θ 2 2 1 sin sin   − θ   θ   = cos 2 θ 3cot 2 θ = 1 cot θ = 1 3 θ = 60 0 383. If 1 cotθ = 3 , 0 0 < θ 0 < 90 0 , then the value of 2 2 2 2-sin θ + cosec θ - secθ 1-cos θ is: (a) 1 (b) 2 (c) 5 (d) 0 SSC CPO-SI 25/11/2020 (Shift-II) Ans. (a) : 1 Cot 3 θ= 0 Cot Cot 60 ⇒ θ= 0 60 ⇒θ= ( ) 2 2 2 2 0 0 2 2 2-Sin θ 2-Sin 60º + Cosec θ-Secθ = +(Cosec 60 -Sec60 ) 1-Cos θ 1-Cos 60º ∴ 2-3/4 4 = + 2 1-1/4 3     −         5/4 2 = + 3/4 3     −         5 2 = - 3 3 3 = 3 = 1 384. Solve for 2 2 1 cos θ - sin θ = 2 0, < θ < 90 0 (a) 60 0 (b) 45 0 (c) 40 0 (d) 30 0 SSC CPO-SI 25/11/2020 (Shift-II) Ans. (d) : Given Cos 2 θ – Sin 2 θ = 1 2 ⇒ Cos2θ = cos60 0 ( ) 2 2 Cos θ-Sin θ = Cos2θ Q ⇒ 2θ = 60 0 ⇒ θ = 30 0 385. If 4 – 2 sin 2 θ – 5 cos θ = 0, 0 o < θ < 90 o , then the value of cosθ – tanθ is: (a) 1 2 3 2 − (b) 1 2 3 2 + (c) 2 3 2 − (d) 2 3 2 + SSC CPO-SI 25/11/2020 (Shift-I) Ans. (a) : 4 – 2sin 2 θ – 5 cosθ = 0 ⇒ 4 – 2 (1–cos 2 θ) – 5 cosθ = 0 ⇒ 4 – 2 + 2 cos 2 θ – 5 cosθ = 0 ⇒ 2cos 2 θ – 5 cos θ + 2 = 0 ⇒ 2cos 2 θ – 4 cos θ – cos θ + 2 = 0 ⇒ 2cosθ (cosθ–2) –1 (cosθ –2) = 0 ⇒ cosθ – 2 = 0 or 2 cosθ – 1 = 0 ⇒ cosθ = 2 Unconsdrable ∴ cos θ = 1 2 ⇒ θ = 60 o ∴ cosθ – tanθ = cos60 o – tan 60 o ⇒ 1 3 2 − ⇒ 1 2 3 2   −     386. If ( ) 2 2 2 0 2 y 4 cosec 57 - tan 33 - cos90 + y × tan 66 tan 24 = 2 , then the value of y is: (a) 4 (b) 8 (c) –4 (d) –8 SSC CPO-SI 25/11/2020 (Shift-I) Ans. (d) : 4 (cosec 2 57 o – tan 2 33 o ) – cos90 o + y×tan 2 66 o × tan 2 24 o = y 2 Q cosec (90 o – θ) = secθ and tan (90 o – θ) = cot θ ⇒ 4 (sec 2 33 o – tan 2 33 o ) – 0 + y × cot 2 24 o × tan 2 24 o = y 2 ⇒ 4 × 1 + y × 1 = y 2 {Q sec 2 θ – tan 2 θ=1} ⇒ y y 4 2 − =− ⇒ y 4 2 =− ⇒ Y = –8 387. The value of 2 2 2 2 sin 52° + 2 + sin 38° 4cos 43° – 5 + 4cos 47° is: (a) –3 (b) 3 (c) 1 3 − (d) 1 3 SSC CPO-SI 24/11/2020 (Shift-II)

Trigonometry 173 . Ans. (a) : 2 2 2 2 sin 52° + 2 + sin 38° 4cos 43° – 5 + 4cos 47° = 2 2 2 2 sin 52 cos 52 2 4sin 47 4cos 47 5 °+ °+ °+ °− ( ) ( ) sin 90 – cos , cos 90 – sin θ= θ     ° θ= θ   Q = 1 2 3 –3 4 1 5 1 + = = ×− − 2 2 sin cos 1   θ+ θ=   Q 388. If 4 (cosec 2 57 0 – tan 2 33 0 ) – cos 90 0 – y tan 2 66 0 tan 2 24 0 y = 2 the value off y is : (a) 8 3 (b) 8 (c) 3 8 (d) 1 3 SSC CPO-SI 23/11/2020 (Shift-I) Ans. (a) ( ) 2 0 2 0 0 2 0 2 0 y 4 cosec 57 - tan 33 -cos90 - ytan 66 .tan 24 = 2 ( ) 2 0 0 2 0 y 4 cosec 90 -33 - tan 33 -0-y= 2     (Q tan A. tan B = 1, if A + B = 90 0 ) 2 0 2 0 y 4 sec 33 tan 33 y 2   ⇒ × − − =   ( ) 2 2 y 4 1 y sec tan 1 2 ⇒ ×− = θ− θ= Q y 4 y 2 ⇒ − = ⇒ 8–2y = y 8 y 3 ⇒ = 389. If 2 2 sin - 3sin +2 =1 cos φ φ φ φ φ φ φ φ φ where 0 0 < φ < 90 0 , then what is the value of (cos2φ + sin3φ + cosec2φ) ? (a) 9 4 3 6 + (b) 3 4 3 6 + (c) 2 3 3 + (d) 3 2 3 3 + SSC CGL (Tier-II) 12-09-2019 Ans. (a) 2 2 sin 3sin 2 1 cos φ− φ+ = φ ( )( ) ( ) 2 sin 1 sin 2 1 1 sin φ− φ− = − φ ( )( ) ( )( ) sin 1 sin 2 1 1 sin sin 1 φ− φ− =− + φ φ− ( ) ( ) sin 2 1 1 sin φ− =− + φ sin 2 sin 1 φ− =− φ− 1 2sin 1 sin 2 φ= ⇒ φ= φ = 30 0 ∴ cos2φ+sin3φ+cosec2φ = cos60 0 +sin90 0 +cosec60 0 1 2 1 2 3 = + + 3 2 2 3 = + 33 4 9 4 3 6 2 3 + + = = 390. If sinθ 1 + cosθ 4 + = 1 + cosθ sinθ 3 ,0 0 < θ < 90 0 , then the value of(tanθ + secθ) –1 is : (a) 3 2 − (b) 2 3 + (c) 2 3 − (d) 3 2 + SSC CGL (Tier-II) 13-09-2019 Ans. (c): sin 1 cos 4 1 cos sin 3 θ + θ + = + θ θ ( ) 2 2 sin 1 cos 2cos 4 sin 1 cos 3 θ+ + θ+ θ = θ + θ ( ) 2 2 cos 4 sin 1 cos 3 + θ = θ + θ 2 4 sin 3 = θ 3 sin 2 θ= θ = 60 0 ( ) ( ) 1 1 0 0 tan 60 sec 60 3 2 − − + = + 1 3 2 3 2 2 3 1 3 2 3 2 − − = × = = − − + − 391. If (A+B+C) = 90 0 , then what is the value of sin (A/2) sin [(180º – B – C)/2] + cos (A/2) sin (B+C)/2 ? (a) 1/2 (b) 1 (c) 1/ 2 (d) 3/2 SSC CGL (Tier-II) 9-3-2018 Ans. (c) : 0 A 180 B C A B C sin .sin cos .sin 2 2 2 2   − − +   +         0 A B C A B C sin .sin 90 cos .sin 2 2 2 2 + +     = − +         A B C A B C sin .cos cos .sin 2 2 2 2 + +   = +    

Trigonometry 174 . A B C sin 2 2 +   = +     [Qsin (A+B)= sinA.cosB+cosA.sinB] 0 90 1 sin 2 2 = = 392. What is the value of cot (90º–x) sin 4 (90º–x) + cot (180º–x) sin 4 (180º–x) ? (a) (cos 4x)/4 (b) (sin 2 2x)/2 (c) (cos 2 2x)/2 (d) (sin 4x)/4 SSC CGL (Tier-II) 9-3-2018 Ans. (d) : cot(90 0 –x).sin 4 (90 0 –x) + cot (180 0 –x).sin 4 (180 0 –x) = tanx.cos 4 x – cotx. sin 4 x = sinx.cos 3 x – cosx.sin 3 x ( ) 2 2 1 2sinx.cosx cos x sin x 2 = × − [Qsin2A=2sinA.cosA] 1 sin 4x sin 2x cos2x 2 4 = × × = 393. The value of sec 2 28 o – cot 2 62 o + sin 2 60 o + cosec 2 30 o is equal to: (a) 3 (b) 23 4 (c) 19 4 (d) 7 2 SSC CGL (TIER-I) – 11.06.2019 (Shift-I) Ans. (b) : sec 2 28 o – cot 2 62 o + sin 2 60 o + cosec 2 30 o = sec 2 28 o – cot 2 (90 0 –28 0 ) + 3 4 4 + = (sec 2 28 o – tan 2 28 0 ) + 19 4 ( ) 2 2 sec tan 1 θ− θ= Q = 19 23 1 4 4 + = 394. sec 2 29 0 – cot 2 61 0 + sin 2 60 0 + cosec 2 30 0 is equal to: (a) 19 4 (b) 23 4 (c) 11 4 (d) 15 4 SSC CGL (TIER-I)– 12.06.2019 (Shift-III) Ans. (b) : 2 0 2 0 2 0 2 0 sec 29 cot 61 sin 60 cosec 30 − + + ( ) ( ) 2 2 2 0 2 0 3 1 tan 29 cot 61 2 2   = + − + +       ( ) 2 0 0 2 0 3 1 tan 90 61 cot 61 4 4 = + − − + + ( ) { } 0 cot 90 tan −θ = θ Q 2 0 2 0 3 1 cot 61 cot 61 4 4 = + − + + 3 1 4 4 = + + 23 4 = 395. The value of sin 2 20 o + sin 2 70 o – tan 2 45 o + sec60 o is equal to: (a) 2 (b) 2.5 (c) 1 (d) 3 SSC CGL (TIER-I) – 12.06.2019 (Shift-II) Ans. (a) : sin 2 20 o + sin 2 70 o – tan 2 45 o + sec60 o = sin 2 20 o + sin 2 (90 0 –20 0 ) – (1) 2 + 2 = (sin 2 20 0 + cos 2 20 0 ) – 1 + 2 = 1 – 1 + 2 = 2 396. The value of 2 2 2 sin 24º sin 66º 2 sin 61º cos 61º sin 29º is : 2 cos 24º cos 66º + + + +       (a) 3 (b) 1 (c) 2 (d) 0 SSC CGL (TIER-I) – 13.06.2019 (Shift-III) Ans. (c) 2 2 2 2 2 sin 24 sin 66 sin 61 cos 61 sin 29 cos 24 cos 66 °+ ° + °+ ° ° °+ ° ( ) ( ) ( ) 2 2 2 2 2 sin 90 66 sin 66 sin 61 cos61 sin 90 61 cos 24 cos 90 24 °− ° + °     = + °+ ° °− ° °+ °− °     ( ) 2 2 2 2 2 cos 66 sin 66 sin 61 cos61 .cos61 cos 24 sin 24 °+ ° = + °+ ° ° °+ ° { } 2 2 1 1 2 sin cos 1 1 = + = θ+ θ= Q 397. If 3 cos 2 A + 7 sin 2 A = 3, 0 0 ≤ A ≤ 90 0 then value of A : (a) 30 0 (b) 0 0 (c) 90 0 (d) 45 0 SSC CGL (TIER-I) – 19.06.2019 (Shift-III) Ans. (b) : 3cos 2 A + 7 sin 2 A = 3 3cos 2 A + 3 sin 2 A + 4sin 2 A = 3 3 × 1 + 4 sin 2 A = 3 ( ) 2 2 sin A cos A 1 + = 4 sin 2 A = 0 ⇒ sin 2 A = 0 sin 2 A = sin 2 0 0 A = 0 0 398. If A + B = 45°, then the value of 2(1+ tanA) (1+ tanB) is: (a) 4 (b) 1 (c) 2 (d) 0 SSC CGL (Tier-I)– 03/03/2020 (Shift-I) Ans. (a) : A + B = 45 0 tan (A+B) = tan45 0 tan A tan B 1 1 tan A. tan B + = − tanA + tanB + tanA. tan B= 1 By add 1 in both side 1 + tanA + tan B + tan A. tan B = 2 (1+tanA) (1+tanB) = 2 2 (1+tanA) (1+tanB) = 4

Trigonometry 175 . 399. sin cos 1 ? sin cos 1 θ− θ+ θ− θ+ θ− θ+ θ− θ+ = θ+ θ− θ+ θ− θ+ θ− θ+ θ− (a) secθ + tanθ (b) secθ – tanθ (c) secθ sinθ (d) secθ tanθ SSC CGL (TIER-I) – 06.06.2019 (Shift-I) Ans. (a) : sin cos 1 sin cos 1 θ− θ+ θ+ θ− By putting the value of θ such that the value of denominator is not zero. If θ = 45 0 1 1 1 1 2 1 2 2 2 1 1 1 2 1 2 1 1 2 2 − + + = = × = + − + + − Solving by the option from option (a), 0 0 sec 45 tan 45 + 2 1 = + 400. The value of (tan29 0 cot61 0 – cosec 2 61 0 ) + cot 2 54 0 – sec 2 36 0 + (sin 2 1 0 + sin 2 3 0 + sin 2 5 0 + ....+ sin 2 89 0 ) is : (a) 21 (b) 1 20 2 (c) 22 (d) 1 22 2 SSC CGL (Tier-II) 12-09-2019 Ans. (b) : (tan29 0 cot61 0 – cosec 2 61 0 ) + cot 2 54 0 – sec 2 36 0 + (sin 2 1 0 + sin 2 3 0 + sin 2 5 0 +....+ sin 2 89 0 ) (cot 2 61 0 – cosec 2 61 0 ) + (cot 2 54 0 – cosec 2 54 0 ) = –1 –1 = – 2 And sin 2 1 0 + sin 2 3 0 + sin 2 5 0 + ...... sin 2 45 0 +...... + sin 2 87 0 + sin 2 89 0 ) if A + B = 90 0 then sin 2 A + sin 2 B = 1 Total term = 45 = (22 ×1) + sin 2 45 0 1 1 22 22 2 2 = + = Required Value = 1 2 22 2 − + 1 20 2 = 401. The value of sin 2 64 0 + cos64 0 sin26 0 + 2cos43 0 cosec47 0 is : (a) 4 (b) 2 (c) 1 (d) 3 SSC CGL (Tier-II) 13-09-2019 Ans. (d) : sin 2 64 0 + cos64 0 .sin26 0 + 2cos43 0 .cosec47 0 = sin 2 64 0 + cos64.cos64 0 + 2cos43 0 .sec43 0 = sin 2 64 0 + cos 2 64 0 + 2 × 1 = 1 + 2 = 3 402. What is the value of sin (90 0 +2A) [4 – cos 2 (90 0 - 2A)]? (a) 4(cos 3 A–sin 3 A) (b) 4(cos 3 A+sin 3 A) (c) 4(cos 6 A+sin 6 A) (d) 4(cos 6 A–sin 6 A) SSC CGL (Tier-II) 19-02-2018 Ans. (d) : 0 2 sin(90 2A) 4 cos (90º 2A)   + − −   2 cos2A(4 sin 2A) = − 2 2 2 2 (cos A sin A) (4 4sin A.cos A) = − − ( )( ) 2 2 2 2 4 cos A sin A 1 sin A.cos A = − − ( )( ) 2 2 2 2 2 2 2 4 cos A sin A sin A. cos A sin A.cos A   = − + −     ( ) 2 2 4 4 2 2 4 cos A sin A (sin A cos A sin A.cos A) = − + + ( ) ( ) 3 3 2 2 4 cos A sin A   = −     ( ) 3 3 2 2 a b a b (a b ab)   − ⇒ − + +   Q 6 6 4 cos A sin A   = −   403. What is the value of ( ) ( )     4 2 1 1 + sin 90 - θ cos 90 - θ -1 ? (a) tan 2 θ sec 2 θ (b) sec 4 θ (c) tan 4 θ (d) tan 2 θ sin 2 θ SSC CGL (Tier-II) 20-02-2018 Ans. (a) : ( ) ( ) 4 2 1 1 sin 90º cos 90º 1 + −θ −θ − 4 2 1 1 cos sin 1 = + θ θ− 4 2 1 1 cos cos = − θ θ 2 4 1 cos cos − θ = θ = 2 4 sin cos θ θ = tan 2 θ.sec 2 θ 404. If 4 – 2 sin 2 θ – 5 cosθ = 0, 0 0 < θ < 90 0 , then the value of sinθ + tanθ is : (a) 2 3 (b) 3 2 (c) 3 2 2 (d) 33 2 SSC CGL (TIER-I) – 04.06.2019 (Shift-I) Ans. (d) : 2 4 2sin θ 5cosθ 0 − − = ( ) 2 4 21 cos θ 5cosθ 0 − − − = 2 4 2 2cos θ 5cosθ 0 − + − = 2 2cos θ 5cosθ 2 0 − + = 2 2cos θ 4cosθ cos θ 2 0 − − + = ( ) ( ) 2cosθ cosθ 2 1 cosθ 2 0 − − − = ( )( ) cos θ 2 2cosθ 1 0 − − = 1 cos θ 2, 2 ∴ = cos θ 2 = Q (Dose not exist)

Trigonometry 176 . 0 1 cos cos 60 2 ∴ θ= = ⇒ θ = 60 0 then, sinθ + tan θ = ? = sin60 0 + tan60 0 3 3 2 = + 33 2 = 405. If sin (x + y) = cos (x – y), then the value of cos 2 x is: (a) 1 2 (b) 3 (c) 1 4 (d) 5 SSC CGL (Tier-II) – 18/11/2020 Ans. (a) : Q sin (x + y) = cos (x – y) x + y + x – y = 90° x 45 = ° cos 2 x = cos 2 45° 2 1 1 2 2   = =     406. If cosec39° = x, then the value of 2 2 1 + sin 39° + cosec 51° 2 2 2 1 tan 51° - sin 51°sec 39° is: (a) 1 – x 2 (b) 1 − 2 x (c) x 2 – 1 (d) 2 x − 1 SSC CGL (Tier-II) – 18/11/2020 Ans. (c) : Cosec 39° = x ⇒ 2 2 2 2 2 1 1 sin 39 tan 51 cosec 51 sin 51º.sec 39 + °+ °− ° ° ⇒ sin 2 51° + cos 2 51° + cot 2 39° – 2 2 1 sin 51.cosec 51° ⇒ 1 + cot 2 39° – 1 ⇒ cot 2 39° ⇒ (cosec 2 39°–1) ⇒ x 2 – 1 407. What is the value of 2 2 2 2 cos 20º +cos 70º - tan 45º? sin 90º (a) 0 (b) 1 (c) –2 (d) –1 SSC CHSL 06/08/2021 (Shift-III) Ans. (a) : 2 2 2 2 cos 20º cos 70º tan 45º sin 90º + − 2 2 2 2 cos 20º sin 20º tan 45º sin 90º + = − ( ) ( ) cos 90 sin −θ = θ Q 1 1 0 1 = − = 408. Simplify the following expression. cos 2 30º + cos 2 40º + cos 2 50º + cos 2 60º (a) 1 (b) 5 2 (c) 3 2 (d) 2 SSC CHSL 16/04/2021 (Shift-III) Ans.(d) : cos 2 30º + cos 2 40º + cos 2 50 + cos 2 60º ( ) ( ) 2 2 2 2 cos 30º cos 90 30º cos 40 cos 90 40º = + − + + − ( ) ( ) 2 2 2 2 cos 30º sin 30º cos 40º sin 40º = + + + ( ) 2 2 sin cos 1 θ+ θ= Q ∴ 1 1 = + =2 409. The value of 2 2 2 2 3cos 27° - 5 + 3cos 63° + sin35°cos55° + cos35°sin55° tan 32° + 4 - cosec 58° is: (a) 1 3 − (b) 2 1 3 (c) 1 3 (d) 1 4 − SSC CHSL 19/08/2021 (Shift-II) Ans. (c) : 2 2 2 2 3cos 27º 5 3cos 63º sin 35º cos 55º cos 35º sin 55º tan 32º 4 cosec 58º −+ + + +− ( ) 2 2 2 2 3cos 27º 3cos 90 27º 5 sin 35º cos 55º cos 35º sin 55º tan 32º 4 cosec 58º + − − = + +− ( ) ( ) 2 2 2 2 3cos 27º 3sin 27º 5 sin(35 55) cot 58º cos ec 58 4 + − = + + − + ( ) 2 2 3 5 sin 90º cosec cot 1 1 4 − = + θ− θ= −+ Q 2 1 3 −   = +     1 3 = 410. The value of       2 2 2 2 2 sin 27º +sin 63º - sin 69º -cos69º sin21º cos 24º +cos 66º is (a) 3 (b) 2 (c) 0 (d) 1 SSC CHSL 11/08/2021 (Shift-I) Ans. (c) : 2 2 2 2 sin 27º +sin 63º cos 24º +cos 66º –sin 2 69º–cos69º sin21º Q sin (90º–θ) = cosθ, cos (90º–θ) = sinθ = 2 2 2 2 sin 27º +sin (90º 27º ) cos 24º +cos (90º 24º ) - - – sin 2 69º–cos69º sin (90º–69º) = 2 2 2 2 sin 27º +cos 27° cos 24º +sin 24° – (sin 2 69º+cos 2 69º) {Q sin 2 θ + cos 2 θ=1} = 1–1 = 0

Trigonometry 177 . 411. ( ) ( )       2 2 2 sinθ 1 + cosθ sec θ + cosec θ 1 + cosθ - sin θ , 0º < θ < 90º is equal to: (a) sec 2 θ (b) cosec 2 θ (c) cotθ (d) tanθ SSC CGL–(Tier-I) 17/08/2021 (Shift I) Ans. (a) : ( ) ( ) 2 2 2 sinθ 1+ cosθ sec θ + cosec θ 1 + cosθ - sin θ       ( ) sinθ 1+ cosθ 1 = sin cos cosθ(1+ cosθ)     θ θ   2 2 2 2 sec cosec sec .cosec   θ+ θ   = θ θ     Q 1 sinθ = sin cos cosθ × θ θ 2 = sec θ 412. The value of (sin 37 0 cos 53 0 + cos 37 0 sin53 0 ) 2 0 2 0 2 0 2 0 4cos 37 - 7 + 4cos 53 – tan 47 + 4 - cosec 43 (a) –2 (b) 0 (c) 1 (d) 2 SSC CGL–(Tier-I) 18/08/2021 (Shift I) Ans. (d) (sin37 0 . cos53 0 + cos37 0 . sin53 0 ) 2 0 2 0 2 0 2 0 4cos 37 7 4cos 53 tan 47 4 cosec 43 − + − + − ( ) 2 0 2 0 0 0 2 0 2 0 4cos 37 7 4sin 37 sin 37 53 tan 47 4 1 sec 47   − + = + −   + − −   0 2 0 2 0 4 7 sin 90 tan 47 4 1 tan 47 − = − + − − 0 4 7 sin 90 3 −   = −     = 1– (–1) = 1 + 1 = 2 413. The value of ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 cosec 26º -tan 64º + cot 42º -sec 48º cot 22º -θ - cosec 62º +θ - tan θ + 68º + tan 28º -θ is: (a) 3 (b) 4 (c) –1 (d) –2 SSC CGL (Tier-II) 03/02/2022 Ans : (d) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 cosec 26 tan 64 cot 42º sec 48º cot 22º cosec 62º tan 68º tan 28º − + − −θ − +θ − θ+ + −θ 2 2 2 2 2 2 3(cosec (90º –64º ) – (tan 64º ) + (cot 42º sec (90º –42º )) cot(90º –(22º –θ)) – cosec (90º –(62º +θ)) – tan(θ + 68º ) + tan (28º –θº ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 sec 64º tan 64º cot 42º cosec 42º tan 68º sec 28º tan 68º tan 28º − + − θ+ − −θ − θ+ + −θ () 31 1 2 2 1 1 − = =− − − 414. The expression (tanθ + cotθ) (secθ + tanθ) (1- sinθ), 0º < θ < 90º, is equal to: (a) secθ (b) cosecθ (c) cotθ (d) sinθ SSC CGL (Tier-II) 03/02/2022 Ans : (b) (tanθ + cotθ) (secθ + tanθ) (1–sinθ) On putting the value of θ = 30º ( )( )( ) tan 30º cot 30º sec30º tan 30º 1 sin 30º = + + − 1 2 1 1 3 1 2 3 3 3     = + + −          4 3 1 2 3 3 = × × = 2 From option (b) Cosecθ = cosec 30º = 2 415. The value of ( )( ) 2 2 2 2 cos9º +sin81º sec9º +cosec81º cosec 71º +cos 15º -tan 19º +cos 75º is: (a) 1 (b) 4 (c) –3 (d) 2 SSC CGL (Tier-II) 03/02/2022 Ans : (d) ( )( ) 2 2 2 2 cos 9º sin 81º sec9º + cosec81º cosec 71º cos 15º tan 19º cos 75º + + − + ( )( ) 2 2 2 2 sin 81º sin 81º cosec81º cosec81º sec 29º sin 75º tan 19º cos 75º + + = + − + 4sin 81º.cosec81º 1 1 = + 4 2 2 = 416. If sin 2 θ–cos 2 θ –3sinθ + 2 = 0, 0° < θ < 90°, then what is the value of 1 + secθ + tanθ? (a) –1+ 3 (b) –1– 3 (c) 1+ 3 (d) 1– 3 SSC CGL (Tier-I) 13/04/2022 (Shift-III) Ans : (c) Given, sin 2 θ – cos 2 θ – 3sinθ + 2 = 0, 0º < θ < 90º Put θ = 30º sin 2 30 – cos 2 30 – 3sin30 + 2 = 0 1 3 3 2 0 4 4 2 − − + = = – 2 + 2 = 0 0 = 0 Value of 1 + secθ + tanθ = ? 2 1 1 3 3 = + + 3 1 3 = + 1 3 = +

Trigonometry 178 . 417. (sinθ + cosθ) 2 = 2, 0° < θ < 90°, then the value of θ is: (a) π (b) 0 (c) 2 π (d) 4 π SSC CHSL –17/03/2020 (Shift-II) Ans. (d) : (sinθ + cosθ) 2 = 2 By putting the value 4 π θ= 2 2 1 1 sin cos 4 4 2 2 π π     + = +         2 2 2   =     4 2 = = 2 Hence the value of θ is 4 π . 418. If sin (A + B) = cos (A-B) = 3 2 and A and B are acute angles. The measure of angle A and B (in degrees) will be: (a) A = 45 and B = 45 (b) A = 45 and B = 15 (c) A = 60 and B = 30 (d) A = 15 and B = 45 SSC CHSL –20/10/2020 (Shift-II) Ans : (b) sin (A + B) = cos (A-B) = 3 2 sin (A + B) = 3 2 = sin 60° A + B = 60° ....(i) cos(A–B) = 3 2 = 30° A – B = 30° ....(ii) From equation (i) and (ii) A = 45°, B = 15° 419. If sin (A – B) = 1/2 and cos(A+B) = 1/2, where A>B>0º and A + B is an acute angle, then the value of A is (a) 60º (b) 30º (c) 45º (d) 15º SSC CHSL –21/10/2020 (Shift-II) Ans. (c) Given, sin (A – B) = 1/2 sin (A – B) = sin30º (A – B) = 30º ...(i) cos (A + B) = 1/2 cos (A + B) = cos60º A + B = 60º ...(ii) On adding the equation (i) and (ii) A – B = 30º A + B = 60º 2A = 90º A = 45º 420. If sin 7x = cos11x, 0º < x < 90º, then the value of tan 9x is: (a) 3 2 (b) 1 (c) 3 (d) 1 3 SSC CHSL –21/10/2020 (Shift-I) Ans. (b) sin 7x = cos11x sin 7x = sin(90 0 – 11x) Qsin(90º–θ) = cosθ 7x = 90 0 – 11x 18x = 90 0 x = 5 0 ∴ tan9x = tan9×5 = tan45º = 1 421. If 3–2 sin 2 θ–3 cosθ = 0, 0 0 < θ < 90 0 then find the value of (2 cosecθ + tan θ) ? (a) 5 3 (b) 5 3 3 (c) 7 3 (d) 7 3 3 SSC CHSL 02/07/2019 (Shift-III) Ans. (d) : 2 3 2sin 3cos 0 − θ− θ= ( ) 2 21 cos 3cos 3 − − θ− θ=− 2 2 2 cos 3cos 3 − + θ− θ=− 2 2 cos 3cos 1 0 θ− θ+ = 2 2 cos 2 cos cos 1 0 θ− θ− θ+ = ( ) ( ) 2 cos cos 1 1 cos 1 0 θ θ− − θ− = ( )( ) 2 cos 1 cos 1 0 θ− θ− = 2 cos 1 and cos 1 θ= θ= 0 cos 1/2 0 θ= θ= 0 60 θ= ( ) 2 cosec + tan =? θ θ On putting θ = 60 0 , 2 2 3 3   = × +     4 3 3   = +     4 3 3 +   =     7 3   =     By rationalization of 3 7 3 3   =      

Trigonometry 179 . 422. If 2 cos 2 θ – 5 cos θ + 2 = 0, 0 0 < θ < 90 0 , then find the value of (cosecθ + cotθ) (a) 1 3 (b) 3 (c) 1 3 (d) 2 3 SSC CHSL 01/07/2019 (Shift-III) Ans. (b) : 2cos 2 θ – 5 cosθ + 2 = 0, 0 0 < θ < 90 0 Formula, 2 b b 4ac x 2a − ± − = ( ) ( ) 2 5 5 4 2 2 cos 2 2 −− ± − − × × θ= × 5 3 cos 4 ± θ= 0 1 cos cos60 2 θ= = By taking the (–) sign. θ = 60 0 then, cosecθ + cotθ = cosec 60 0 + cot60 0 2 1 3 3 = + 3 3 = = 3 423. Write the exact value of the given below : 2 0 2 0 2 0 0 0 2 0 2 0 sin 25 sin 65 sin 71 cos71 sin19 cos 24 cos 66         + + + + + + + + +         +         (a) 3 (b) 0 (c) 1 (d) 2 SSC CHSL 02/07/2019 (Shift-II) Ans. (d) : 2 0 2 0 2 0 0 0 2 0 2 0 sin 25 sin 65 sin 71 cos71 sin19 cos 24 cos 66   + + +   +   ( ) ( ) ( ) 2 0 2 0 0 2 0 0 2 0 2 0 0 0 0 sin 25 sin 90 25 sin 71 cos 71 cos 24 cos 90 24 .sin 90 71   + −   + + + −   =     −   2 0 2 0 2 0 0 0 2 0 2 0 sin 25 cos 25 sin 71 cos 71 .cos 71 cos 24 sin 24   + = + +   +   2 0 2 0 1 sin 71 cos 71   = + +   1 1 2 = + = 424. If tanθ + cotθ = 2 and θ is acute, then the value of tan 100 θ + cot 100 θ is equal to: (a) 2 (b) 3 (c) 1 (d) 0 SSC CHSL –26/10/2020 (Shift-I) Ans. (a) : tan θ + cot θ = 2 Let θ = 45 0 tan 45 0 + cot 45 0 = 2 1 + 1 = 2 2 = 2 As per question 100 100 tan cot ? θ+ θ= 100 0 100 0 tan 45 cot 45 ? + = 1 + 1 = 2 425. If ( ) ( ) ( ) ( ) 1 + sinθ – cosθ 1 + sinθ + cosθ + =4 1 + sinθ + cosθ 1 + sinθ – cosθ , then which of the following values will be suitable for θ ? (a) 30 0 (b) 90 0 (c) 45 0 (d) 60 0 SSC CHSL –13/10/2020 (Shift-II) Ans. (a) : ( ) ( ) ( ) ( ) 1 sin – cos 1 sin + cos 4 1 sin + cos 1 sin – cos + θ θ + θ θ + = + θ θ + θ θ = ( ) ( ) 2 2 2 2 21 sin 2sin + cos 1 sin cos + θ+ θ θ + θ − θ = 4 ( ) ( ) ( ) 2 2 22 2sin 4 1 sin 1 sin + θ = = + θ − − θ = ( ) ( ) 41 sin 4 21 sin sin + θ = + θ θ ⇒ sinθ = 1 2 = sin30 0 ⇒ θ = 30 0 426. Find x if cosx = − 1 2 . (a) 4 3 π (b) 2 3 π (c) 5 3 π (d) 3 2 π SSC CHSL–14/10/2020 (Shift-III) Ans. (b) : 1 cos x 2 − = cos x = cos (180 0 – 60 0 ) cos x = cos 120 0 x = 120 0 = 2 3 π 427. If + θ θ 1 1 cosec +1 cosec -1 = 2secθ, 0º < θ < 90º , then the value of θ θ θ tan + 2sec cosec (a) 2 2 2 + (b) 2 3 2 + (c) 4 3 2 + (d) 4 2 2 + SSC CPO-SI – 09/12/2019 (Shift-II)

Trigonometry 180 . Ans. (d) 1 1 cos ec 1 cos ec 1 + θ+ θ− = 2secθ 2 cos ec 1 cos ec 1 cos ec 1 θ− + θ+ θ− = 2secθ 2 2cosec cot θ θ = 2 1 c osθ 2 2 1 sin 2 sin cos θ ⋅ × θ θ = 2. 1 c osθ tanθ = 1 = tan45 0 θ = 45 0 ∴ tan 2sec cos ec θ+ θ θ = 0 0 0 tan 45 2sec45 cos ec45 + = 1 2 2 2 2 2 + × = 4 2 2 + 428. If sec 3x = cosec (3x–45 o ), where 3x is an acute angle, then x is equal to: (a) 45 o (b) 22.5 o (c) 35 o (d) 27.5 o SSC CPO-SI 23/11/2020 (Shift-II) Ans. (b) : Sec 3x = cosec (3x–45 o ) sec 3x = sec [90 o – (3x – 45 o )] 3x = 135° – 3x 6x = 135° ⇒ x = o 135 22.5 6 ° = 429. If 4θ is an acute angle, and cot 4θ = tan(θ–5°), then what is the value of θ? (a) 24° (b) 45° (c) 21° (d) 19° SSC CPO-SI 24/11/2020 (Shift-II) Ans. (d) : cot 4θ = tan (θ – 5 0 ) tan (90° – 4θ) = tan(θ – 5 0 ) 95° = 5θ θ = 19° 430. Find the value of 2 0 2 0 2 0 2 0 cos 30 - sin 30 sin 15 + cos 15 . (a) 1 2 (b) 0 (c) 1 3 − (d) 1 SSC CHSL –15/10/2020 (Shift-III) Ans. (a) : 2 0 2 0 2 0 2 0 cos 30 -sin 30 sin 15 + cos 15 = 2 2 2 2 3 1 2 2 [ sin cos 1] 1     −           θ+ θ= Q = 3 1 4 − = 1 2 431. Find the value of 2 o 2 o 2 o 2 o 2 o o o o o 1 1 4tan 30 + sin 90 + cot 60 +sin 30 cos 45 4 8 sin60 cos30 - cos60 sin30 (a) 1 2 2 (b) 4 (c) 3 1 4 (d) 1 3 2 SSC CHSL (Tier-I) 03/07/2019 (Shift-III) Ans. (d) : 2 o 2 o 2 o 2 o 2 o o o o o 1 1 4tan 30 + sin 90 + cot 60 +sin 30 cos 45 4 8 sin60 .cos30 - cos60 .sin30 = () 2 2 2 2 2 1 1 1 1 1 1 4 + 1 + + . 4 8 2 3 3 2 3 3 11 . . 2 2 22                         − = 4 1 1 1 + + 3 4 24 8 3 1 4 4 + − = 32 6 1 3 + + 24 24 24 24 2 4 + = 42 4 24 2 × = 21 7 6 2 = = 1 3 2 432. The value of tan 2 48º – cosec 2 42º + cosec (67º + θ) – sec (23º – θ) is: (a) –2 (b) 0 (c) –1 (d) 1 SSC CPO-SI 13/12/2019 (Shift-I) Ans. (c) tan 2 48 0 – coses 2 42 0 + cosec (67 0 + θ) – sec(23 0 –θ) = tan 2 48 0 – sec 2 48 0 + sec(23 0 – θ) – sec(23 0 – θ) = –1 433. If 2 sin (60º – α) = 1, 0 0 < α < 90º, then α is equal to: (a) 15 0 (b) 30 0 (c) 45 0 (d) 60 0 SSC CPO-SI – 12/12/2019 (Shift-II) Ans. (a) 2 sin (60 0 – α) = 1 sin (60 0 – α) = 1 2 sin (60 0 – α) = sin 45 0 60 0 – α = 45 0 α = 15 0

Trigonometry 181 . 434. If 0 ≤ θ ≤ 90 0 , and sin (2θ + 50 0 ) = cos (4θ + 16 0 ), then what is the value of θ (in degrees)? (a) 10 0 (b) 4 0 (c) 8 0 (d) 12 0 SSC CPO-SI – 09/12/2019 (Shift-I) Ans. (b) sin(2θ + 50 0 ) = cos(4θ + 16 0 ) sin(2θ + 50 0 ) = sin (90 0 – (4θ + 16 0 )) 2θ + 50 0 = 90 0 – 4θ – 16 0 6θ = 74 0 – 50 0 θ = 4 0 435. The value of 2 0 2 0 sin 38 sin 52 + + + + + + + + 2 0 sin 30 2 0 tan 45 − is equal to: (a) 1 2 (b) 3 4 (c) 1 4 (d) 1 3 SSC CGL (TIER-I)– 10.06.2019 (Shift-II) Ans. (c) : 2 0 2 0 2 0 2 0 sin 38 sin 52 sin 30 tan 45 + + − ( ) 2 0 2 0 0 2 0 2 0 sin 38 sin 90 38 sin 30 tan 45 = + − + − 2 0 2 0 2 0 2 0 sin 38 cos 38 sin 30 tan 45 = + + − 2 1 1 1 2   = + −     1 4 = 436. If tan 4θ = cot (2θ + 30 o ), then θ is equal to: (a) 25 o (b) 15 o (c) 20 o (d) 10 o SSC CGL (TIER-I) – 11.06.2019 (Shift-I) Ans. (d) : tan 4θ = cot (2θ + 30 o ) tan 4θ = tan [90 0 – (2θ + 30 o )] 4θ = 90 o – (2θ + 30 o ) 6θ = 60 o θ = 10 o 437. If cosec 4θ = sec (60 0 – 2θ) then θ is equal to: (a) 20 0 (b) 15 0 (c) 25 0 (d) 18 0 SSC CGL (TIER-I)-2018 – 12.06.2019 (Shift-III) Ans. (b) : cosec4θ = sec(60 0 – 2θ) ( ) 0 1 1 sin 4 cos 60 2 = θ − θ ( ) 0 cos 60 2 sin 4 − θ= θ ( ) ( ) 0 0 cos 60 2 cos 90 4 − θ= − θ 0 0 60 2 90 4 ∴ − θ= − θ 0 2 30 θ= 0 15 θ= 438. The value of 2 o 2 o o o o o sin 30 + cos 60 - sec35 .sin55 sec60 + cosec 30 is equal to (a) 1 8 (b) 1 4 − (c) 1 4 (d) 1 8 − SSC CPO-SI 23/11/2020 (Shift-II) Ans. (d) : 2 o 2 o o o o o sin 30 + cos 60 - sec35 .sin55 sec60 + cosec 30 = 2 2 o o 1 1 1 sec35 2 2 sec 35 2 2     + −         + 1 1 1 1 1 4 4 2 4 4 + − − = = = 1 8 − 439. The value of 2 o 2 o o o o o sin 30 + cos 60 + sec45 .sin45 sec60 + cosec30 is: (a) 3 8 − (b) 1 4 (c) 1 4 − (d) 3 8 SSC CPO-SI 24/11/2020 (Shift-I) Ans. (d) : 2 o 2 o o o o o sin 30 + cos 60 + sec45 .sin45 sec60 + cosec30 = 2 2 1 1 1 2. 2 2 2 2 2     + +         + = 1 1 1 4 4 4 + + = 6 1 4 4 × = 3 8 440. If sin 3x = cos (3x – 45 O ), 0 O < 3x < 90 O , then x is equal to: (a) 22.5 O (b) 45 O (c) 35 O (d) 27.5 O SSC CPO-SI 24/11/2020 (Shift-I) Ans. (a) : sin 3x = cos (3x – 45 O ), 0 O < 3x < 90 O If sin A = Cos B then A + B will be 90º. ∴ 3x + (3x – 45 O ) = 90 O 6x = 135 O x = 135 6 x = 22.5 O 441. Solve the following. 0 0 0 0 0 0 0 2sin22 2cot75 8tan45 tan20 tan40 tan50 tan70 - - 0 0 5 cos68 5tan15 ? (a) 2 (b) 1 (c) 3 (d) 0 SSC CGL (Tier-I)– 09/03/2020 (Shift-II)

Trigonometry 182 . Ans. (d) : 0 0 0 0 0 0 0 0 0 2 sin 22 2 cot 75 8 tan 45 . tan 20 . tan 40 . tan 50 . tan 70 5 cos 68 5tan15 − − 0 0 0 0 0 0 0 0 2 cos 68 2 cot 75 8 1 tan 20 .tan 40 .cot 40 .cot 20 5 cos 68 5cot 75 ×× = − − 2 8 2 2 2 0 5 5 = − − = − = 442. Solve the following: sin0 0 sin30 0 sin45 0 sin60 0 sin90 0 = ? (a) 1 (b) 4 (c) 0 (d) 6 8 SSC CGL (Tier-I) – 06/03/2020 (Shift-II) Ans. (c) : sin0 0 .sin30 0 .sin45 0 .sin60 0 .sin90 0 [Q sin0 0 = 0] = 0 443. Solve the following. 0 0 0 0 0 0 sin40 cosec50 + - 4cos50 cosec40 cos50 sec40 (a) 1 (b) –2 (c) 2 (d) –1 SSC CGL (Tier-I) – 06/03/2020 (Shift-I) Ans. (b) 0 0 0 0 0 0 sin40 cosec50 + - 4cos50 cosec40 cos50 sec40 = 0 0 0 0 0 0 cos 50 sec 40 4 sin 40 .cos ec40 cos 50 sec 40 + − × 0 0 sin(90 ) cos cos ec(90 ) sec   −θ = θ   −θ = θ     Q = 1 + 1 – 4 × 1 = – 2 444. Find the value of 0 0 0 0 2 0 2 0 2 0 tan30 cosec60 + tan60 sec30 sin 30 + 4cot 45 - sec 60 ? (a) 2 3 (b) 32 99 (c) 8 3 (d) 32 3 SSC CGL (Tier-I)– 04/03/2020 (Shift-I) Ans. (d) : 0 0 0 0 2 0 2 0 2 0 tan30 cosec60 + tan60 sec30 sin 30 + 4cot 45 -sec 60 1 2 2 3 3 3 3 1 4 4 4 × + × = + − 2 2 8 4 32 3 1 3 3 4 + × = = = 445. The value of cos0°cos30°cos45°cos60°cos90° is: (a) 3 (b) 6 8 (c) 0 (d) 5 SSC CGL (Tier-I) – 05/03/2020 (Shift-III) Ans. (c) : cos0°cos30°cos45°cos60°cos90° = 0 [Q cos 90° = 0] 446. If sin3θ = cos(20 0 – θ), then θ is equal to: (a) 25 (b) 28 (c) 35 (d) 30 SSC CGL (TIER-I) – 13.06.2019 (Shift-I) Ans. (c) : As per question, ( ) 0 sin 3 cos 20 θ= −θ ( ) ( ) 0 0 cos 90 3 cos 20 −θ= −θ ( ) { } 0 cos 90 x sin x − = Q 0 0 90 3 20 ∴ − θ= −θ 0 0 3 90 20 θ−θ= − 0 2 70 θ= 0 35 θ= 447. The value of 0 0 0 tan30 + tan60 cos30 is: (a) 8 3 (b) 3 3 + (c) 8 3 (d) 1 3 + SSC CGL (Tier-I)– 03/03/2020 (Shift-III) Ans. (a) : 0 0 0 tan 30 tan 60 cos 30 + 1 3 3 3 2 + = 4 8 3 3 3 2 = = 448. If 2sinθ – 8 cos 2 θ + 5 = 0, 0° < θ < 90°, then what is the value of (tan2θ + cosec2θ)? (a) 5 3 3 (b) 33 (c) 4 3 3 (d) 2 3 SSC CGL (Tier-I)– 03/03/2020 (Shift-III) Ans. (a) : By value putting, θ = 30 0 2sin30 0 – 8cos 2 30 0 + 5 1 3 2 8 5 0 2 4 = × − × + =

Trigonometry 183 . 0 0 tan 2 cosec2 tan 60 cos ec60 ∴ θ+ θ= + 2 5 5 3 3 3 3 3 = + = = 449. The value of the expression cosec(85°+ θ) – sec(5° – θ) – tan(55° + θ) + cot(35°–θ) is: (a) –1 (b) 3 2 (c) 0 (d) 1 SSC CGL (Tier-I) – 03/03/2020 (Shift-II) Ans. (c) : cosec (85 0 +θ) – sec(5 0 –θ)–tan(55 0 +θ) + cot(35 0 –θ) = cosec [90 0 – (5 0 –θ)] – sec (5 0 –θ) – tan [90 0 –(35 0 –θ)] + cot (35 0 –θ) = sec (5 0 –θ) – sec (5 0 –θ) – cot (35 0 –θ) + cot (35 0 –θ) = 0 450. The value of ( ) ( ) ( ) 0 0 2 0 2 0 0 0 0 0 sin 78 +θ -cos 12 -θ + tan 70 -cosec 20 sin25 cos65 + cos25 sin65 is : (a) –1 (b) –2 (c) 2 (d) 0 SSC CGL (Tier-II) 12-09-2019 Ans. (a) ( ) ( ) ( ) 0 0 2 0 2 0 0 0 0 0 sin 78 cos 12 tan 70 cosec 20 sin 25 cos65 cos 25 sin 65 +θ − −θ + − + ( ) ( ) ( ) ( ) 0 0 2 0 2 0 0 0 sin 78 sin 78 cot 20 cosec 20 sin 25 65 +θ − +θ + − = + [Q cosec 2 θ – cot 2 θ = 1] sin (90º–θ) = cosθ] 0 0 1 1 1 sin 90 1 − − = = =− 451. The value of 0 0 0 0 0 0 0 0 (cos 9 + sin 81 )(sec 9 + cosec 81 ) sin 56 sec 34 + cos 25 cosec 65 is : (a) 4 (b) 1 2 (c) 2 (d) 1 4 SSC CGL (Tier-II) 11-9-2019 Ans. (c) : 0 0 0 0 0 0 0 0 (cos 9 + sin 81 )(sec 9 + cosec 81 ) sin 56 sec 34 + cos 25 cosec 65 ( ) ( ) 0 0 0 0 0 0 0 0 0 0 sin 81 sin 81 sec 9 sec 9 4sin81 cosec81 2 2 sin 56 .cosec 56 cos 25 .sec 25 + + × = = = + 452. The value of cosec (67 0 + θ) – sec (23 0 – θ) + cos15 0 cos35 0 cosec55 0 cos60 0 cosec75 0 is : (a) 1 2 (b) 1 (c) 0 (d) 2 SSC CGL (Tier-II) 13-09-2019 Ans. (a) : cosec(67 0 +θ) – sec(23 0 –θ) + cos15 0 .cos35 0 . cosec55 0 .cos60 0 .cosec75 0 = sec(23 0 –θ) – sec (23 0 –θ) + cos15 0 .cos35 0 .sec35 0 × 1 2 × sec15 0 1 1 0 2 2 = + = 453. What is the value of [tan 2 (90º – θ) – sin 2 (90º – θ)] cosec 2 (90º – θ) cot 2 (90º – θ)? (a) 0 (b) 1 (c) –1 (d) 2 SSC CGL (Tier-II) 17-2-2018 Ans. (b) : [tan 2 (90 0 –θ) – sin 2 (90 0 –θ)] cosec 2 (90 0 –θ) cot 2 (90 0 -θ) 2 2 2 2 cot cos sec .tan   = θ− θ θ θ   2 2 2 2 2 1 1 sin cos 1 . sin cos cos θ   = θ −   θ θ θ   2 2 2 2 4 1 sin sin cos . sin cos   − θ θ = θ   θ θ   {Q 1–sin 2 θ = cos 2 θ} = 1 454. What is the value of cot (90 0 + 75 0 ) ? (a) 2 3 + (b) 2 3 − − (c) 3 1 + (d) 3 1 − − SSC CGL (Tier-II) 9-3-2018 Ans. (b) : cot (90º + 75º) = –tan75º = –tan (30º + 45º) tan 30º tan 45º 1 tan 30º . tan 45º +   =−   −   1 1 3 1 3 1 3 1 3 1 3 1 1 1 3   +     + +   =− =− ×     − +     − ×     3 1 2 3 2 3 2   + + =− =− −       455. What is the value of sin (630 0 + A ) + cos A ? (a) 3/2 (b) 1/2 (c) 0 (d) 2/ 3 SSC CGL (Tier-II) 18-02-2018 Ans. (c) : sin(630° + A) + cosA = sin [2×360º–(90º–A)]+ cosA = – sin(90º – A) + cosA = – cosA + cosA = 0 456. What is the value of [cos (90º+A) ÷ sec (270º – A)] + [(sin 270º + A) ÷ cosec (630º–A)]? (a) 3 sec A (b) tan A sec A (c) 0 (d) 1 SSC CGL (Tier-II) 19-02-2018

Trigonometry 184 . Ans. (d) : ] ( ) 0 cos(90 A) sec(270º A) [sin (270º A) cos ec 630º A   = + ÷ − + + ÷ −   [ ] [ ] sin A cos ecA cos A sec A = ÷ + ÷ 2 2 sin A cos A = + = 1 457. Find the value of sin 2 35º + sin 2 55º. (a) 1 (b) 0 (c) -1 (d) 1 2 SSC CHSL –15/10/2020 (Shift-II) Ans. (a) : sin 2 35º + sin 2 55º = sin 2 35º + sin 2 (90 0 – 35º) = sin 2 35º + cos 2 35º = 1 458. If α + β = 90º and α = 2β, then the value of cos 2 α + sin 2 β is? (a) 1 (b) 1/3 (c) 1/5 (d) 1/2 SSC CHSL –21/10/2020 (Shift-III) Ans. (d) Given, α + β = 90º α = 2β Let, α = 60º 60º = 2 × 30º β = 30º 60º = 60º Now the required value of cos 2 α + sin 2 β = ? cos 2 60º + sin 2 30º = ? 2 2 1 1 ? 2 2     + =         1 1 1 4 4 2 + = 459. The value of o o o o o 2 o 2 o cos29 cosec61 tan45 + 2sin35 sec55 3sin 42 + 3sin 48 is: (a) 3 (b) 1 (c) 0 (d) 2 SSC CHSL –19/03/2020 (Shift-III) Ans. (b) : o o o o o 2 o 2 o cos29 cosec61 tan45 + 2sin35 sec55 3sin 42 + 3sin 48 = ( ) o o o o 2 2 o o cos 29 1 1 2sin35 sin 61 cos 55 3sin 42 3 sin 90 42 ×+   + −     = ( ) ( ) o o o o o o 2 o 2 o cos 29 2 sin 35 sin 90 29 cos 90 35 3sin 42 3cos 42 + − − + = ( ) o o o o 2 o 2 o cos 29 2 sin 35 cos 29 sin 35 3 sin 42 cos 42 + + = 1 2 1 3 + = 460. Find the value of Cos 225°. (a) 0.7071 (b) –0.866 (c) 0.866 (d) –0.7071 SSC CHSL –17/03/2020 (Shift-II) Ans. (d) : Cos 225º = Cos(180º + 45º) = – Cos45º 1 – 2 = = – 0.7071 461. 0 0 0 0 0 0 0 0 0 0 1 - tanA tan3 tan15 tan30 tan75 tan87 = 1 + tanA tan27 tan39 tan51 tan60 tan63 , then the value of cot A is: (a) 1 (b) 3 (c) 4 (d) 2 SSC CHSL –18/03/2020 (Shift-I) Ans. (d) : o o o o o o o o o o 1 tan A tan 3 tan15 tan 30 tan75 tan87 1 tan A tan 27 tan 39 tan 51 tan 60 tan 63 − = + o o o o o o o o o o tan 3 tan15 tan 30 cot15 cot 3 tan 27 tan 39 cot 39 tan 60 cot 27 = = o o tan 30 1/ 3 1 3 3 tan 60 = = ∴ 1 tan A 1 1 tan A 3 − = + By componendo dividendo rule 2 4 2 tan A 2 = − − 1 2 tan A = ⇒ cot A = 2 462. If x = tan 40 o , then the value of 2 tan 50° will be: (a) 2 x (b) 2x (c) 1 x (d) 1 2x SSC CHSL –19/03/2020 (Shift-II) Ans. (a) : Q x = tan 40 o ∴ 2 tan50 o = 2 tan (90 o – 40 o ) = 2 cot 40 o = o 2 tan 40 ⇒ 2 x 463. The value of 0 0 0 0 0 0 sin30 cos60 + cos45 sin45 tan60 cot30 is: (a) 1 3 (b) 2 (c) 3 (d) 1 4 SSC CHSL –19/03/2020 (Shift-I)

Trigonometry 185 . Ans. (d) : o o o o o o sin 30 cos60 cos 45 sin 45 tan 60 cot 30 + = 1 1 1 1 2 2 2 2 3 3 × + × × = 1 1 4 2 3 + ⇒ 3/4 3 ⇒ 3 1 4 3 × ⇒ 1 4 464. If A = 60°, then what is the value of (4cos 3 A– 3cosA)? (a) 3 2 − (b) –1 (c) 1 (d) 3 2 SSC CHSL 16/04/2021 (Shift-I) Ans. (b) : 4cos 3 A–3cosA = cos3A (by formula) cos3A = cos180° by putting the value of A = 60° cos180º = –1 465. If tan3θ = sin45° . cos45° + cos60° and 3θ is an acute angle, then what will be the value of sin4θ? (a) 1/2 (b) 3 /2 (c) 1/ 2 (d) 1 SSC CHSL 12/08/2021 (Shift-I) Ans. (b) : tan3θ = sin45° . cos45° + cos60° (Given) tan3θ = 1 1 2 2 1 × + 2 = 1 tan3θ = 1 = tan45° 3θ = 45° θ = 15° ∴ sin4θ = sin60° = 3 2 466. What is the value of sin33°cos57° + sec62°sin28° + cos33°sin57° + cosec62°cos28° tan15°tan35°tan60°tan55°tan75° ? (a) 3 (b) 3 3 (c) 2 3 (d) 2 SSC CHSL 16/04/2021 (Shift-I) Ans. (a) : sin33°cos57° + sec62°sin28° + cos33°sin57° + cosec62°cos28° tan15°tan35°tan60°tan55°tan75° = sin33°sin33° + cosec28°sin28° + cos33°cos33° + sec28°cos28° cot75°cot55°tan60°tan55°tan75° Q sinθ cosecθ = 1, secθ. cosθ = 1 and tanθ.cotθ = 1 = 2 2 sin 33 1 cos 33 1 11 3 °+ + °+ ×× = 1 1 1 3 3 3 + + = = 3 467. Evaluate the following expression. 2 2 2 2 2 2 tan 60º +cosec30ºsin90º +3sec 30º 4sin 45º +sec 60º -cot 30º -5cos 90º (a) 7 3 (b) 3 (c) 19 17 (d) –12 SSC CHSL 06/08/2021 (Shift-I) Ans. (b) : 2 2 2 2 2 2 tan 60º +cosec30º sin90º +3sec 30º 4sin 45º +sec 60º cot 30º 5cos 90º - - = 4 3 2 1 3 9 3 1 3 4 4 3 5 0 2 + ×+ × = × + − − × = 3 468. If sin θ (2sinθ + 3) = 2, 0° < θ < 90°, then what is the value of (sec 2 θ + cot 2 θ – cos 2 θ)? (a) 13/3 (b) 31/12 (c) 7/2 (d) 43/12 SSC CHSL 19/04/2021 (Shift-I) Ans. (d) : sinθ (2 sinθ + 3) = 2 (0° < θ < 90°) 2sin 2 θ + 3 sinθ = 2 2sin 2 θ + 3 sinθ – 2 = 0 2sin 2 θ + 4 sinθ – sinθ – 2 = 0 (2sinθ–1) (sinθ +2) 2sinθ = 1 sinθ = 1 2 = sin30° θ = 30º sin θ + 2 = 0 sinθ = –2 According to the question, sec 2 θ + cot 2 θ – cos 2 θ = ? = sec 2 30° + cot 2 30° – cos 2 30° = 4 3 3 3 4 + − = 16 36 9 12 + − = 16 27 12 + = 43 12 Trick:– sinθ(2sinθ + 3) = 2 takingθ = 30º L.H.S = sin30º (2sin30º+3) ( ) 1 4 2 2 = × =

Trigonometry 186 . L.H.S = R.H.S ∴sec 2 θ + cot 2 – cos 2 θ ∵ θ =30º 4 3 3 3 4 = + − 43 12 = 469. If θ is an acute angle and sinθ = cosθ, then the value of 2 tan 2 θ + sin 2 θ – 1 is equal to: (a) 3 (b) 1 (c) 3/2 (d) –7 SSC CHSL 15/04/2021 (Shift-I) Ans. (c) : sin θ = cos θ (Qθ = Acute angle) tanθ = 1 tanθ = tan45º θ = 45° According to the question, 2tan 2 θ + sin 2 θ –1 = 2tan 2 45° + sin 2 45° –1 = 2×1+ 1 2 –1 = 3 2 470. If 0 0 3tanθ = 2 3sinθ, 0 < θ < 90 , then find the value of 2sin 2 2θ –3cos 2 θ 3θ. (a) 3 2 − (b) 1 2 (c) 3 2 (d) 1 SSC CGL–(Tier-I) 13/08/2021 (Shift III) Ans. (c) : Given that– 3 tan 2 3 sin ,0º 90º θ= θ ∠θ∠ 3sin 2 3 sin cos θ ⇒ = θ θ 3 2 3 cos ⇒ = θ 1 2 3 cos 3 3 ⇒ = θ × 1 2 cos 3 ⇒ = θ cos cos30º ⇒ θ= ∴θ = 30º ∴ 2sin 2 2θ – 3cos 2 3θ = 2sin 2 (2 × 30º) – 3cos 2 (3×30º) = 2sin 2 60º – 3cos 2 90º 2 3 2 3 0 2   = × − ×       3 2 4 = × 2 2 3 2sin 2 3cos 3 2 θ− θ= 471. If sin 2 θ – cos 2 θ – 3 sinθ + 2 = 0, 0 0 < θ < 90 0 , then what is the value of 1 secθ - tanθ is? (a) 4 3 (b) 2 2 (c) 2 3 (d) 4 2 SSC CGL–(Tier-I) 2308/2021 (Shift I) Ans. (a) : Given that, sin 2 θ – cos 2 θ – 3sinθ + 2 = 0 sin 2 θ – 1 + sin 2 θ – 3 sinθ + 2 = 0 2sin 2 θ – 3sinθ + 1 = 0 sin θ = 1 and sinθ 1 2 = Hence θ = 30 0 [Q0º< θ < 90º] 4 1 1 3 3 sec30º tan 30º 2 1 3 3 ∴ = = = − − OR We can solve this question by value putting method on putting θ = 30 0 472. If 2cos 2 θ = 3sinθ, 0 0 < θ < 90 0 , then the value of (sec 2 θ – tan 2 θ + cos 2 θ) is : (a) 9 4 (b) 5 4 (c) 3 4 (d) 7 4 SSC CGL–(Tier-I) 13/08/2021 (Shift II) Ans. (d) : Given, 2cos 2 θ = 3sinθ, 0º< θ<90º Put, θ = 30º 3 1 2 3 4 2 × = × 3 3 2 2 = (Sec 2 θ – tan 2 θ + cos 2 θ) [Q sec 2 – tan 2 θ = 1] ∴1 + cos 2 30 3 1 4   = +     7 4 = 473. If 3 cosθ = , 2 then the value of 2 2 2 2-sin θ + (sec θ + cosecθ) 1-cot θ is: (a) 25 12 (b) 59 24 (c) – 25 12 (d) – 59 24 SSC CHSL 05/08/2021 (Shift-I) Ans. (b) : cos θ = 3 2 --------------[Given] cos θ = cos 30° θ = 30° 2 2 2 2 sin θ + (sec θ + cosecθ) 1 cot θ - -

Trigonometry 187 . = ( ) 2 2 2 2 2 + +2 3 1             1 - 2 -3 = 1 2 4 4 2 1 3 3 − + + − = – 7 4 2 4 2 3 + + × = 21 32 48 24 − + + = 59 24 474. If tan x = cot (48º + 2x), and 0º < x < 90º, then what is the value of x? (a) 12º (b) 14º (c) 16º (d) 21º SSC CHSL 05/08/2021 (Shift-I) Ans. (b) : tan x = cot (48° + 2x) (0°<x<90°) tan x = cot (2x+48°) cot (90°–x) = cot (2x+48°) On comparing the both side 90°–48° = 2x + x 3x = 42° x = 14° 475. If cos53º x = y , then sec 53º + cot 37º is equal to: (a) 2 2 x y x y + − (b) 2 2 x y x x + − (c) 2 2 y y x x + − (d) 2 2 y y x y + − SSC CGL (Tier-I) 19/04/2022 (Shift-II) Ans. (c) x cos 53º y = 2 sin 53º 1 cos 53º = − 2 2 x 1 y = − 2 2 y x /y = − ∴sec53º + cot 37º = sec53º + tan 53º 1 sin 53º cos 53º cos53º = + 1 sin 53º cos 53º + = 2 2 1 y x /y x/y + − = 2 2 y y x x + − = 476. If cosec A = secB, where A and B are acute angles, then what is the value of (A + B)? (a) 0º (b) 135º (c) 90º (d) 145º SSC CGL (Tier-I) 21/04/2022 (Shift-II) Ans : (c) Given, cosecA = sec B 1 sec B sin A = if A B 90 sin A.sec B 1 + = °     =   Q 1 = sin A .secB then A+B = 90° 477. If tan 2 x-3tanx + 2 = 0 and (0º < x <90º) , then the value of x is: (a) 90º (b) 30º (c) 45º (d) 60º SSC CHSL 15/04/2021 (Shift-III) Ans.(c) : tan 2 x – 3tanx + 2 = 0 tan 2 x – 2tanx – tanx + 2 = 0 tanx (tanx–2) –1 (tanx – 2) = 0 (tanx–2) (tanx–1) = 0 tanx = 2(is not value of any angle) ∴tan x = 1 = tan45º x = 45º 478. Solve for θ : 3cosecθ+4sinθ – 4 3 = 0, where θ is an acute angle. (a) 30° (b) 45° (c) 15° (d) 60° SSC CHSL 11/082021 (Shift-II) Ans. (d) : 3cosecθ + 4sinθ – 4 3 = 0 Let, θ= 60º L.H.S = 3cosecθ + 4sinθ– 4 3 = 3× cosec60º + 4 sin60º – 4 3 2 3 3 4 4 3 2 3 = × + × − 2 3 2 3 4 3 0 R.H.S = + − = = L.H.S = R.H.S Hence, θ=60º 479. The value of 2 tan50º +sec50º + cos 65º +sin65º cos25º +tan30º cot40º +cosec40º is: (a) ( ) 3 3 1 3 + (b) 2 + √3 (c) 6 3 3 + (d) 1 + √3 SSC CHSL 05/08/2021 (Shift-I) Ans. (c) : tan 50° + sec 50° cot 40° + cosec 40° + cos 2 65°+sin65°cos25°+tan30°

Trigonometry 188 . = tan(90° 40°) + sec(90° 40°) cot 40 cosec 40 °+ ° - - + cos 2 65°+sin65° × cos(90°–65°)+tan30° = 1+cos 2 65° + sin 2 65° + 1 3 = 2 + 1 3 = 2 3 1 3 6 3 3 3 3 + + × = 480. If α + β = 90° and , α = 2β, then the value of 2 2 3cos α - 2sin β is equal to : (a) 4 3 (b) 3 4 (c) 3 2 (d) 1 4 SSC CGL (Tier-II)-2019 – 18/11/2020 Ans. (d) : Q α + β = 90° α = 2β (given) 3β = 90° 2β + β = 90° β = 30° And α = 60° then, 3 cos 2 α – 2sin 2 β = ? 3 cos 2 60º – 2 sin 2 30° 2 2 1 1 3 2 2 2     × − ×         3 2 1 4 4 − = 481. The value of 2 0 0 0 tan 60 + sin90 - 2tan45 is (a) 0 (b) 4 (c) 1 (d) 2 SSC CGL (Tier-I) – 05/03/2020 Ans. (a) : 2 0 0 0 tan 60 sin 90 2 tan 45 + − = 3 1 2 1 + − × = 2 – 2 = 0 482. If x, y are acute angles, where 0 < x + y < 90 0 and sin (3x – 40 0 ) = cos (3y + 40 0 ), then the value of tan (x + y) is equal to: (a) 1 2 (b) 1 3 (c) 3 (d) 1 3 SSC CHSL –26/10/2020 (Shift-III) Ans. (b): sin (3x – 40 0 ) = cos (3y + 40 0 ) if sin θ = cos α then θ + α = 90 0 ∴ 3x – 40 0 + 3y + 40 0 = 90 0 3 (x+y) = 90 0 x + y = 30 0 Now tan (x+y) = tan30 0 = 1 3 483. The value of 0 0 0 0 0 0 0 0 sin23 cos67 tan45 + cos23 sin67 cot45 2sin45 cos45 is: (a) 0 (b) 2 (c) 1 (d) 1 2 SSC CHSL –26/10/2020 (Shift-II) Ans. (c) : sin 23 cos 67 tan 45 cos 23 sin 67 cot 45 2sin 45 cos 45 ° ° °+ ° ° ° ° ° sin 23 .cos 67 cos 23 sin 67 1 ° °+ ° ° = = sin (23° + 67°) = sin 90° = 1 484. If cos 27 0 = x, then the value of tan 63 0 is: (a) 2 1 x x + (b) 2 x 1 x + (c) 2 1 x x − (d) 2 x 1 x − SSC CHSL –26/10/2020 (Shift-III) Ans. (d) Given : cos 27° = x 1 (AB) 2 = (AC) 2 + (BC) 2 (1) 2 = (AC) 2 + (x) 2 1 – x 2 = (AC) 2 2 1 x AC − = Then, tan63 0 = 2 x 1 x − 485. If sin(2x-45 0 ) = cosx and angle x and (2x–45 0 ) (in degrees) are acute angles, then the value of cotx is: (a) 0 (b) 1 2 (c) 1 (d) 1 2 SSC CHSL –26/10/2020 (Shift-II) Ans. (c) : sin (2x – 45 0 ) = cos x cos [90 0 – (2x – 45 0 )] = cos x cos (90 0 – 2x + 45 0 ) = cos x 135 0 = 3x x = 45 0 Then, cot x = cot 45 0 =1

Trigonometry 189 . 486. The value of sin 60 o cos 30 o – cos 60 o sin30 o is: (a) 1 2 (b) 3 2 (c) 1 2 (d) 1 SSC CHSL –19/10/2020 (Shift-II) Ans. (c) : o o o o sin 60 cos 30 cos 60 sin 30 − Q ( ) sin A B sin A cos B cos A sin B − = − ∴ ( ) o o o o o o sin 60 cos 30 cos60 sin 30 sin 60 30 − = − = o 1 sin 30 2 = 487. What is the value of o o o o o cot35 cot40 cot45 cot50 cot55 ? (a) 1 (b) 2 (c) 0 (d) -1 SSC CHSL –16/10/2020 (Shift-III) Ans. (a) : o o o o o cot35 cot40 cot45 cot50 cot55 = o o o o o o cot35 cot40 .cot 45° cot(90 - 40 ). cot (90 -35 ) = o o o o cot 35 cot 40 ×1× tan 40 tan 35 = o o o o tan 35 cot 35 . tan 40 cot 40 1 tan cot   θ=   θ   Q = 1×1 = 1 488. Which of the following values for A to make the equation 0 0 0 0 0 A tan62 sec28 cot38 =1 cosec62 tan11 is true? (a) 0 0 0 tan 38 tan 79 tan28 (b) 0 0 0 tan 38 tan79 tan28 (c) 0 0 0 tan 28 tan79 tan38 (d) 0 0 0 tan 28 tan38 tan79 SSC CHSL –14/10/2020 (Shift-II) Ans. (d) : 0 o 0 0 o A tan 62 .sec 28 .cot 38 1 cosec62 .tan11 = 0 o 0 0 o Acot 28 .cosec 62 .cot38 1 cosec 62 .cot 79 = A = 0 0 0 cot 79 cot28 .cot38 = 0 0 0 tan28 .tan38 tan 79 (V) Miscellaneous 489. Find the value of the following. 2+ 2+ 2 + 2cos8θ (a) sin 2 θ (b) 2 cos θ (c) cos 2 θ (d) 2 cos 2 θ SSC CGL (Tier-I) 27/07/2023 (Shift-III) Ans. (b) : 2+ 2+ 2 + 2cos8θ ( ) 2 2 21 cos8 + + + θ ( ) 2 2 2 21 2cos 4 1 = + + + θ− 2 [ cos 2θ = 2cos θ – 1] Q = 2 2 2 2 2cos 4 + + × θ 2 2 2 cos 4 = + + θ ( ) 2 21 cos 4 = + + θ 2 2cos2 = + θ (Similarly) 2 2(1+2cos θ–1) = 2cosθ = 490. Study the given triangle and find the length of BC. (a) 6 (b) 3 (c) 5 2 (d) 5 SSC CGL 03/12/2022 (Shift-IV) Ans. (d) : sinθ = P H x sin 30 10 °= 1 x 2 10 x 5 ⇒ = ∴ = 491. If A = 2(sin 6 θ + cos 6 θ) – 3(sin 4 θ + cos 4 θ) then find the value of 3α while cos α = 3+A 5+A (a) 135 o (b) 45 o (c) 180 o (d) 90 o SSC MTS 13/09/2023 (Shift I st ) Ans. (a) : A = 2 (sin 6 θ + cos 6 θ) – 3 (sin 4 θ + cos 4 θ) A = 2 (1 – 3 sin 2 θ.cos 2 θ) – 3 (1 – 2 sin 2 θ.cos 2 θ) A = 2 – 6sin 2 θ.cos 2 θ – 3 + 6 sin 2 θ.cos 2 θ A = –1 3 A 1 cos 5 A 2 + α= = + cos cos 45 α= ° α = 45° 3α = 3 × 45° = 135°

Trigonometry 190 . 492. What is the value of 0 0 0 0 0 0 cos1 .cos2 cos3 ...cos177 .cos178 .cos179 ? (a) 1 (b) 1 2 (c) 1 2 (d) 0 SSC Selection Posts XI-28/06/2023 (Shift-III) Ans. (d) : ⇒ cos1 0 .cos2 0 .cos3 0 ............cos177 0 .cos178 0 .cos179 0 and ⇒ cos1 0 .cos2 0 .cos3 0 ......cos90 0 .cos177 0 .cos178 0 .cos179 0 (จ cos90 0 = 0) ⇒ 0 493. If 3 sin x + 4 cos x = 2, then the value of 3 cos x – 4 sin x is equal to: (a) 23 (b) 29 (c) 21 (d) 21 SSC CGL (Tier-II)- 18/11/2020 Ans. (d) : Q 3 sin x + 4 cos x = 2 By squaring the both side. (3 sin x + 4 cos x) 2 = (2) 2 9 sin 2 x + 16 cos 2 x + 24 sin x. cos x = 4 9 (1–cos 2 x) + 16 (1– sin 2 x) + 24 sin x. cos x = 4 9 – 9 cos 2 x + 16 – 16 sin 2 x + 24 sin x. cos x = 4 9 + 16 – 4 = 9 cos 2 x + 16 sin 2 x – 24 sin x. cos x 21 = (3 cos x – 4 sin x) 2 Hence 3 cos x – 4 sin x = 21 494. What is the value of {(sin 4x + sin 4y) [(tan (2x – 2y)]}/(sin 4x–sin 4y) ? (a) tan 2 (2x + 2y) (b) tan 2 (c) cot (x–y) (d) tan (2x + 2y) SSC CGL (Tier-II) 21-02-2018 Ans. (d): ( ) ( ) { } ( ) sin 4x sin 4y tan 2x 2y / sin 4x sin 4y + − −     ( ) ( ) 4x 4y 4x 4y 2sin .cos sin 2x 2y 2 2 4x 4y 4x 4y cos 2x 2y 2cos .sin 2 2 + −         −     = × + − −             ( ) ( ) ( ) ( ) ( ) ( ) sin 2x 2y .cos 2x 2y sin 2x 2y cos 2x 2y .sin 2x 2y cos 2x 2y + − − = × + − − ( ) tan 2x 2y = + 495. What is the value of [cos 3θ + 2cos5θ + cos 7θ) ÷ (cos θ+ 2 cos 3θ + cos5θ)] + sin 2θ tan 3θ? (a) cos 2θ (b) sin 2θ (c) tan 2θ (d) cot θ sin 2θ SSC CGL (Tier-II) 19-02-2018 Ans. (a) : ( ) cos3 2cos5 cos 7 sin 2 .tan3 cos 2cos3 cos 5 θ+ θ+ θ + θ θ θ+ θ+ θ cos3 cos7 2cos5 sin2 .tan3 cos cos5 2cos3 θ+ θ+ θ = + θ θ θ+ θ+ θ 2cos5 .cos2 2cos5 sin 2 .tan3 2cos3 .cos2 2cos3 θ θ+ θ = + θ θ θ θ+ θ ( ) ( ) 2cos5 cos 2 1 sin 3 sin 2 2cos3 cos 2 1 cos 3 θ θ+ θ = + θ× θ θ+ θ cos5 sin2 .sin3 cos3 θ+ θ θ = θ ( ) cos 2 3 sin 2 .sin 3 cos 3 θ+ θ + θ θ = θ cos2 .cos3 cos 3 θ θ = θ = cos 2θ 496. What is the value of [(tan 5θ + tan 3θ)/4 cos 4θ (tan 5θ – tan 3θ)]? (a) sin 2θ (b) cos 2θ (c) tan 4θ (d) cot 2θ SSC CGL (Tier-II) 17-2-2018 Ans. (b) : ( ) ( ) tan5 tan 3 4cos4 tan5 tan 3 θ+ θ = θ θ− θ sin 5 sin 3 cos5 cos 3 sin 5 sin 3 4.cos4 cos5 cos 3 θ θ + θ θ = θ θ   θ −   θ θ   ( ) ( ) ( ) ( ) sin5 .cos3 cos5 .sin3 sin 5 3 cos5 .cos3 4.cos4 sin5 .cos3 cos5 .sin3 4.cos4 sin 5 3 cos5 .cos3 θ θ+ θ θ θ+ θ θ θ = = θ θ θ− θ θ θ θ− θ θ θ sin8 2sin4 .cos4 2sin2 .cos2 cos 2 4cos4 .sin2 4cos4 .sin2 2.sin 2 θ θ θ θ θ = = = = θ θ θ θ θ θ 497. What is the value of [(sin 7x–sin 5x) ÷ (cos 7x + cos 5x)] – [(cos 6x–cos 4x) ÷ (sin 6x + sin 4x)] ? (a) 1 (b) 2 tan x (c) tan 2 x (d) tan (3 x/2) SSC CGL (Tier-II) 18-02-2018 Ans. (b) : Given– sin 7x sin 5x cos 6x cos 4x cos 7x cos 5x sin 6x sin 4x − −   ⇒ −   + +   7x 5x 7x 5x 6x 4x 6x 4x 2cos .sin sin .sin 2 2 2 2 7x 5x 7x 5x 6x 4x 6x 4x 2cos cos 2sin cos 2 2 2 2 + −  + −                              ⇒ − + − + −                               = tan x + tan x = 2 tan x 498. What is the value of [sin (y–z) + sin (y+z) + 2 sin y]/[sin (x–z) + sin (x+z) + 2 sin x] ? (a) cos x sin y (b) (sin y)/(sin x) (c) sin z (d) sin x tan y SSC CGL (Tier-II) 20-02-2018

Trigonometry 191 . Ans. (b) : ( ) ( ) ( ) ( ) sin y z sin y z 2siny sin x z sin x z 2sinx − + + +     − + + +     y z y z y z y z 2sin cos 2sin y 2 2 x z x z x z x z 2sin cos 2sin x 2 2  − + + − − −      +             =  − + + − − −      +             ( ) ( ) 2 sinycos z sin y 2 sin x. cos z sin x − +     = − +     ( ) ( ) sin y cos z 1 sin x cos z 1 − +     = − +     sin y sin x = 499. What is the value of {[sin (x+y) – 2 sin x + sin (x–y)]/[cos (x–y) + cos (x+y) – 2 cosx]} × [(sin 10x–sin 8x)/(cos 10x + cos 8x)] ? (a) 0 (b) tan 2 x (c) 1 (d) 2 tan x SSC CGL (Tier-II) 20-02-2018 Ans. (b): ( ) ( ) ( ) ( ) ( ) ( ) sin x y 2 sin x sin x y sin10x sin 8x cos10x cos8x cos x y cos x y 2 cos x   + − + −   −     ×   + − + + −         [ ] [ ] 10x 8x 10x 8x 2 cos .sin 2sin x.cos y 2 sin x 2 2 10x 8x 10x 8x 2 cos x.cos y 2 cos x 2cos .cos 2 2 + −         −     = × + − −             ( ) ( ) 2 sin x cos y 1 2 cos 9x. sin x 2 cos x cos y 1 2 cos 9x. cos x − = × − 2 2 sin x cos x = = tan 2 x 500. If tan θ + secθ = (x–2)/(x+2), then what is the value of cos θ ? (a) (x 2 – 1)/ (x 2 + 1) (b) (2x 2 – 4)/ (2x 2 + 4) (c) (x 2 – 4)/(x 2 + 4) (d) (x 2 – 2)/(x 2 + 2) SSC CGL (Tier-II) 21-02-2018 Ans. (c) : tanθ + secθ = (x–2)/(x+2) ....... (i) Q sec 2 θ – tan 2 θ = 1 ⇒ (secθ–tanθ) (secθ+tanθ) = 1 ⇒ ( ) 1 sec tan sec tan θ− θ= θ+ θ ⇒ x 2 sec tan .........(ii) x 2 + θ− θ= − ⇒ x 2 x 2 2 sec x 2 x 2 − +     θ= +     + −     ( ) ( ) 2 2 2 x 4x 4 x 4x 4 2 sec x 4 − + + + + θ= − ( ) 2 2 2x 4 2 sec x 4 + ⇒ θ= − 2 2 x 4 cos x 4 − ⇒ θ= + 501. secθ + tanθ secθ – tanθ is equal to: (a) 1 sec tan θ− θ (b) 1 sec tan θ+ θ (c) ( ) 2 sec tan θ+ θ (d) ( ) 2 sec tan θ− θ SSC CHSL –26/10/2020 (Shift-II) Ans. (c) : secθ + tanθ secθ + tanθ secθ – tanθ secθ + tanθ × (By rationalization) ( ) 2 2 2 sec tan sec tan θ+ θ θ− θ (secθ + tanθ) 2 ( 2 2 sec tan 1 θ− θ= Q ) 502. If secθ= 13 5 , then the value of – 10 tanθ + 24cosecθ 39sinθ 10secθ is: (a) 2 (b) 1 5 (c) 3 (d) 5 SSC Sel. Post Phase VIII (G.L.) 09.11.20 (Shift-2) Ans. (d) : Q 13 Hypotenuse sec 5 Base θ= = From triplet, Perpendicular = 12 then, 12 13 10 24 10 tan 24cosec 5 12 12 13 39sin 10sec 39 10 13 5 × + × θ+ θ = θ− θ × − × 24 26 50 5 36 26 10 + = = = − 503. If 1 + tanθ = 3 , then 3 cotθ – 1 = _______. (a) 3 1 2 + (b) 3 1 2 − (c) 2 3 1 2 − (d) 2 3 1 2 + SSC CHSL 03/06/2022 (Shift- II) Ans. (a) : 1 + tanθ = 3 tanθ = 3 – 1 1 3 1 cot = − θ cotθ 1 3 1 = − On multiplying by 3 + 1 in numerator and denominator cotθ 1 3 1 3 1 3 1 + = × − + cotθ = 3 1 2 + ............... (i)

Trigonometry 192 . 3 cotθ – 1 Putting value from eq n (i) = 3 1 3 1 2 + × − 3 3 2 2 + − = 3 1 2 + = 504. If cosec 2 θ + cot 2 θ 1 , 3 = where 0 ≤ θ ≤ , 2 π then the value of cosec 4 θ – cot 4 θ is: (a) 2 3 (b) 1 3 − (c) 1 3 (d) 2 3 − SSC CHSL 09/06/2022 (Shift- I) Ans. (c) : cosec 2 θ + cot 2 θ 1 (Given) 3 = ∴ cosec 4 θ – cot 4 θ = (cosec 2 θ – cot 2 θ) (cosec 2 θ + cot 2 θ) Q (cosec 2 θ – cot 2 θ) = 1 = 1 1 3 × = 1 3 505. The value of ( ) sin4θ is : 1 - cos4θ (a) cotθ (b) cot2θ (c) tanθ (d) tan2θ SSC CHSL 01/06/2022 (Shift- III) Ans. (b) : ( ) sin 4 1 cos 4 θ − θ 2 sin 2 2sin cos 2sin 1 cos 2 θ= θ θ       θ= − θ     Q 2 sin 2(2 ) 2sin2 .cos2 cos 2 cot 2 1 cos2(2 ) sin 2 2sin 2 θ θ θ θ = = = θ − θ θ θ 506. The value of (1 + tan10º) (1 + tan35º) is: (a) 1 2 (b) 3 4 (c) 1 (d) 2 SSC CHSL 25/05/2022 (Shift- III) Ans. (d) : If A + B = 45º Q (1+tanA) (1+tanB)=2 here, ∴ (1 + tan10º) (1 + tan35º) = 2 507. If cot 75º = 2 – 3 . Find the value of cot 15º. (a) 2 – 3 (b) 2 + 3 (c) 3 + 1 (d) 3 – 1 SSC CHSL 24/05/2022 (Shift- III) Ans. (b) : cot75º = 2 – 3 , cot 15º = ? From ∆ABC, 1 cot15º 2 3 = − = 1 2 3 2 3 2 3 + × − + = 2 3 4 3 + − 2 3 = + ⇒ cot 15º = 2 + 3 508. What is the value of [2 cot(π – A)/2]/[1 + tan 2 (2π–A)/2]? (a) 2 sin 2 A/2 (b) cos A (c) sin A (d) 2 cos 2 A/2 SSC CGL (Tier-II) 21-02-2018 Ans. (c) : 2 A 2 cot 2 2 A 1 tan 2 π−       = π−   +     2 A 2 cot 2 2 A 1 tan 2 π   −     =   + π−     2 2 tan A / 2 1 tan A/2 = + sin A = (From the formula) 509. What is the value of [1–sin(90 0 -2A)]/ [1+sin(90 0 +2A)]? (a) sin A cos A (b) cot 2 A (c) tan 2 A (d) sin 2 A cos A SSC CGL (Tier-II) 19-02-2018 Ans. (c) : ( ) ( ) 0 0 1 sin 90 2A 1 sin 90 2A − − + + = 1 cos 2A 1 cos 2A − + ( ) 2 2 1 1 2sin A 1 2cos A 1 − − = + − 2 2 2 2sin A tan A 2cos A = =

Trigonometry 193 . 510. If 2 2 2 cos θ =3 cot θ-cos θ , where 0° < θ < 90° then the value of θ is : (a) 60° (b) 45° (c) 50° (d) 30° SSC CGL (Tier-II)-2019 – 18/11/2020 Ans. (a) : 2 2 2 cos 3 cot cos θ = θ− θ where 0° < θ < 90° On dividing by cos 2 θ in numerator and denominator in left side. 2 1 3 cos ec 1 = θ− 2 1 3 cot = θ tan 2 θ = 3 tanθ = 3 ∴ 60 θ= ° 511. If 3 cot A = 4 tan A and A is an acute angle, then what will be the value of sec A? (a) 7 2 (b) 21 3 (c) 1 3 (d) 1 2 SSC CHSL 06/08/2021 (Shift-III) Ans. (a) : 3cotA = 4tanA. 2 3 tan A 4 = ∵ sec 2 A = 1 + tan 2 A 2 3 7 So, sec A 1 4 4 = + = sec A 72 = 512. If sinθ + cosθ =3 sinθ - cosθ , then the value of sin 4 θ – cos 4 θ is equal to: (a) 2 5 (b) 4 5 (c) 3 5 (d) 1 5 SSC CHSL 19/04/2021 (Shift-III) Ans. (c) : sin cos 3 sin cos θ+ θ = θ− θ 3sinθ – 3cosθ = sinθ + cosθ 4cosθ = 2sinθ tanθ = 2 ∵ sec 2 θ = 1 + tan 2 θ ⇒ sec 2 θ = 1 + (2) 2 = 5 1 cos 5 θ= 2 sin 5 θ= 4 4 16 1 15 3 sin cos 25 25 25 5 ∴ θ− θ= − = = 513. In ∆ABC, if ∠B = 90°, AB = 21cm and BC = 20 cm, then - 1 + sinA cosA 1 + sinA + cosA is equal to: (a) 3 5 (b) 3 5 − (c) 2 5 (d) 2 5 − SSC CHSL 12/04/2021 (Shift-III) Ans : (c) 20 21 1 1 sin A cos A 29 29 20 21 1 sin A cos A 1 29 29 + − + − = + + + + 29 1 29 41 − = + 28 4 2 70 10 5 = = = 514. What is the value of [tan (90º–A) + cot (90º– A)] 2 ÷ [2sec 2 (90º–2A)] ? (a) 0 (b) 1 (c) 2 (d) –1 SSC CGL (Tier-II) 20-02-2018 Ans. (c) : ( ) ( ) ( ) 2 2 tan 90º A cot 90º A 2sec 90º 2A − + −       −   ( ) 2 2 cot A tan A 2cosec 2A   +       2 2 2 2 2 cos A sin A cos A sin A sin A cos A sin A.cos A 2cosec 2A 2cosec 2A   +   +     =                ( ) 2 2 2 sin 2A 2 cos ec 2A = 2 2 4 2A cosec 2 cosec 2A = = 2

Trigonometry 194 . 515. If 2sinθ + 15cos 2 θ = 7, 0° < θ < 90°, then tanθ + cosθ + secθ = (a) 4 (b) 4 3 5 (c) 3 3 5 (d) 3 SSC CGL (Tier-I) – 03/03/2020 (Shift-II) Ans. (c) : 2sinθ + 15cos 2 θ = 7 15 – 15sin 2 θ + 2sinθ = 7 15 sin 2 θ – 2sinθ – 8 = 0 15sin 2 θ – 12 sinθ + 10 sinθ – 8 = 0 3sinθ (5sinθ – 4) + 2(5sinθ – 4) = 0 (5sinθ – 4) (3sinθ + 2) = 0 3sinθ + 2 = 0 ⇒ sinθ = 2 3 − And 5 sinθ – 4 = 0 ⇒ 4 sin 5 θ= ∴ tanθ = 4 3 , cosθ = 3 5 leLee secθ = 5 3 then, tan cos sec θ+ θ+ θ 4 3 5 3 5 3 = + + 3 3 3 3 5 5 = + = 516. If 7 sin 2 θ – cos 2 θ + 2sinθ = 2, 0° < θ < 90°, then the value of sec2θ + cot2θ cosec2θ + tan2θ is: (a) ( ) 2 1 3 5 + (b) ( ) 1 1 2 3 5 + (c) 1 (d) 2 3 1 3 + SSC CGL (Tier-I) – 04/03/2020 (Shift-II) Ans. (b) : 7sin 2 θ – cos 2 θ + 2sinθ = 2 Putting the value of θ = 30 0 , 1 3 1 7 2 2 4 4 2 × − + × = 4 1 2 4 + = 2 = 2 sec 2 cot 2 cosec2 tan 2 θ+ θ ∴ θ+ θ 0 0 0 0 sec60 cot 60 cos ec60 tan 60 + = + 1 2 2 3 1 3 2 5 3 3 + + = = + = ( ) 1 1 2 3 5 = + 517. If 11 sin 2 θ – cos 2 θ + 4 sinθ – 4 = 0, 0° < θ < 90°, then what is the value of cos2θ + cot2θ sec2θ - tan2θ ? (a) 12 7 3 6 + (b) 10 53 3 + (c) 10 7 3 6 + (d) 12 53 3 + SSC CGL (Tier-I)– 05/03/2020 (Shift-II) Ans. (a) : 11Sin 2 θ – Cos 2 θ + 4 sinθ – 4 = 0 Putting the value of θ = 30° , 1 3 1 11 4 4 4 4 2 × − + × − 8 2 4 0 4 + − = ∴ LHS = RHS Hence, 1 1 0 0 cos 60 cot 60 3 2 (2 3) 2 3 0 0 2 3 (2 3)2 3 (2 3) Sec60 tan 60 + + + + = = × − − + − = 7 4 3 3 12 7 3 6 2 3 3 + + × = 518. If 12cos 2 θ – 2sin 2 θ + 3cosθ = 3, 0° < θ < 90°, then what is the value of cosecθ + secθ tanθ + cotθ ? (a) 2 3 4 + (b) 1 2 2 2 + (c) 1 3 2 + (d) 4 3 4 + SSC CGL (Tier-I)– 05/03/2020 (Shift-I) Ans. (c) : 12 cos 2 θ – 2sin 2 θ + 3cosθ = 3 By putting the value θ = 60° 1 3 1 12 2 3 3 4 4 2 × − × + × = 3 = 3 ∴ 0 0 .0 0 2 2 2 3 2 cosec60 sec 60 3 3 1 4 tan 60 cot 60 3 3 3 + + + = = + + = 1 3 2 + 519. If (cos 2 θ–1) (1+tan 2 θ) + 2tan 2 θ= 1, 0 0 ≤ θ ≤ 90 0 then θ is: (a) 90 0 (b) 30 0 (c) 45 0 (d) 60 0 SSC CGL (Tier-I) – 06/03/2020 (Shift-II)

Trigonometry 195 . Ans. (c) : (cos 2 θ – 1) (1+tan 2 θ) + 2tan 2 θ = 1 –sin 2 θ × sec 2 θ + 2tan 2 θ = 1 –tan 2 θ + 2tan 2 θ = 1 tan 2 θ = 1 tan 2 θ = tan 2 45º, θ = 45 0 520. If (2sinA + cosecA) = 2 2 , 0° < A < 90°, then the value of 2(sin 4 A + cos 4 A) is: (a) 2 (b) 1 (c) 4 (d) 0 SSC CGL (Tier-I) – 06/03/2020 (Shift-I) Ans. (b) : 2sinA + cosecA = 2 2 By putting the value A = 45°, 2 2 2 2 2 1 × + = ∴ 2(sin 4 A + cos 4 A) = 2(sin 4 45° + cos 4 45°) = 2 1 1 1 4 4   + =     521. If 5 cos 2 θ + 1 = 3 sin 2 θ, 0 0 < θ < 90 0 , then what is the value of tanθ + secθ cotθ + cosecθ ? (a) 3 2 3 3 + (b) 2 33 2 + (c) 2 33 3 + (d) 3 2 3 2 + SSC CGL (Tier-I)– 07/03/2020 (Shift-III) Ans. (a): 5cos 2 θ + 1 = 3 sin 2 θ 5 cos 2 θ + 5sin 2 θ+1 = 8sin 2 θ (5 × 1) + 1 = 8 sin 2 θ 8sin 2 θ = 6 3 3 sin 4 2 θ= = θ = 60 0 0 0 0 0 tan 60 sec 60 3 2 3 2 1 2 cot 60 cos ec60 3 3 3 + + + ∴ = = + + 3 2 3 3 + = 522. In a ∆ABC, right angled at B, AB = 7cm and (AC–BC) = 1 cm. The value of (sec C + cot A) is: (a) 4 3 (b) 3 4 (c) 19 24 (d) 1 SSC CGL (TIER-I) – 10.06.2019 (Shift-I) Ans. (a) : Right angle ∆ABC AC – BC = 1 (Given) ....... (i) AC 2 = AB 2 + BC 2 AC 2 – BC 2 = AB 2 (AC – BC) (AC + BC) = 7 2 AC + BC = 49 ...........(ii) By solving the equation (i) and (ii). AC = 25, BC = 24 25 7 32 sec C cot A 24 24 24 ∴ + = + = 4 3 = 523. If 0 0 < θ < 90 0 and cos 2 θ = 3 (cot 2 θ – cos 2 θ) then the value of 1 1 sec sin 2 −   θ+ θ     is: (a) ( ) 2 3 1 − (b) 3 1 + (c) 3 2 + (d) ( ) 22 3 − SSC CGL (TIER-I) – 04.06.2019 (Shift-II) Ans. (d) : cos 2 θ = 3cot 2 θ – 3cos 2 θ 4cos 2 θ = 2 2 3cos sin θ θ sin 2 θ = 3 4 sinθ = 3 2 θ = 60 o 1 1 sec sin 2 −   ∴ θ+ θ     1 1 3 2 2 2 −   = × +       1 2 3 2 −   + =       ( ) ( ) ( ) 22 3 2 3 2 3 − = + − ( ) 22 3 = − 524. If 4-2sin 2 θ-5cosθ = 0,0 o < θ < 90 o then the value of cosθ +tanθ is : (a) 1 2 3 2 − (b) 2 3 2 + (c) 1 2 3 2 + (d) 2 3 2 − SSC CPO-SI 23/11/2020 (Shift-I)

Trigonometry 196 . Ans. (c) : 4–2 sin 2 θ – 5 cosθ = 0 ⇒ 4 – 2 (1 –cos 2 θ) – 5cos θ = 0 ⇒ 4 – 2 + 2cos 2 θ – 5 cos θ = 0 ⇒ 2 cos 2 θ – 5 cosθ + 2 = 0 ⇒ 2 cos 2 θ – 4 cosθ – cosθ + 2 = 0 ⇒ 2 cos θ (cos θ – 2) – 1(cos θ – 2) = 0 ⇒ (2 cos θ – 1) (cos θ – 2) = 0 ⇒ 2cos θ –1 = 0 or cos θ – 2 = 0 ⇒ cos θ 1 2 = or cos θ = 2 (Does not exist) [ ] 1 cos 1 − ≤ θ≤ Q ⇒ cos θ = cos 60° ⇒ θ = 60° ∴ cos θ + tan θ = cos 60° + tan 60° 1 3 2 = + 1 2 3 2 + = 525. If tan (11θ) = cot(7θ), then what is the value of sin 2 (6θ) + sec 2 (9θ) + cosec 2 (12θ)? (a) 43 12 (b) 23 6 (c) 31 12 (d) 35 12 SSC CPO-SI – 11/12/2019 (Shift-I) Ans. (a) tan (11θ) = tan(90º – 7θ) 11θ = 90º – 7θ θ = 5 0 ∴ sin 2 30º + sec 2 45º + cosec 2 60º = ( ) 2 2 2 1 2 2 2 3     + +         = 4 2 4 3 1 + + = 43 12 526. If sinθ – cosθ = 0, 0º < θ < 90º, then the value of sin 4 θ + cos 4 θ is: (a) 1 3 (b) 1 (c) 1 2 (d) 1 4 SSC CPO-SI – 12/12/2019 (Shift-II) Ans. (c) sinθ – cosθ = 0 sinθ = cosθ sinθ = sin (90 0 – θ) θ = 90 0 – θ θ = 45 0 Hence sin 4 θ + cos 4 θ = sin 4 45 0 + cos 4 45 0 = 4 4 1 1 2 2     +         = 1 1 4 4 + = 1 2 527. If θ θ θ cos cosθ + = 4, 0º < < 90º 1-sin 1 + sinθ , then what is the value of (secθ + cosecθ + cotθ)? (a) 2 + 3 (b) 2 3 3 + (c) 1 + 2 3 (d) 1 2 3 3 + SSC CPO-SI 13/12/2019 (Shift-II) Ans. (a) cosθ cosθ + 1-sinθ 1+ sinθ = 4 2 cos + cos sin + cos - cos sin 4 1 sin θ θ θ θ θ⋅ θ = − θ 2 2 cos 4 cos θ = θ 0 1 cos cos 60 2 θ= = θ = 60 0 ∴ secθ + cosecθ + cotθ = sec60 0 + cosec60 0 + cot60 0 2 1 2 3 2 1 2 3 3 3 + + + + = = 2 3 + 528. If 0º ≤ θ ≤ 90 0 , and sec 107 θ + cos 107 θ = 2, then, (secθ + cosθ) is equal to: (a) 1/2 (b) 2 (c) 1 (d) 2 –107 SSC CPO-SI – 09/12/2019 (Shift-I) Ans. (b) sec 107 θ + cos 107 θ = 2 0 0 ≤ θ ≤ 90 0 By putting the value of θ = 0 0 1 + 1 = 2 ⇒ 2 = 2 Hence secθ + cosθ = sec0 0 + cos0 0 = 1+1 = 2 529. If 2 2 sin .sec , 3 θ θ= θ θ= θ θ= θ θ= 0 0 < θ < 90 0 then find the value of (tan 2 θ + cos 2 θ) : (a) 11 12 (b) 5 4 (c) 13 12 (d) 7 6 SSC CHSL 03/07/2019 (Shift-I)

Trigonometry 197 . Ans. (c) : 2 2 sin .sec 3 θ θ= sin 2 .sec cos 3 θ ⇒ θ= θ 2 tan .sec 3 ⇒ θ θ= 1 2 tan .sec . 3 3 ⇒ θ θ= Hence it is clear that, 0 0 tan .sec tan 30 .sec30 θ θ= 0 30 ∴ θ= 2 2 2 0 2 0 tan cos tan 30 cos 30 ⇒ θ+ θ= + 2 2 1 3 2 3     = +           1 3 3 4 = + 4 9 13 12 12 + = = 530. If 2 2 2 (sinθ + cosecθ) + (cosθ + secθ) = k + tan θ 2 +cot θ , then the value of k is equal to : (a) 7 (b) 5 (c) 9 (d) 2 SSC CGL (Tier-II) – 18/11/2020 Ans. (a) : (sinθ + cosecθ) 2 + (cosθ + secθ) 2 = k + tan 2 θ + cot 2 θ sin 2 θ + cosec 2 θ + 2sinθ.cosecθ + cos 2 θ + sec 2 θ + 2cosθ.secθ = k + tan 2 θ + cot 2 θ (sin 2 θ+cos 2 θ)+(cosec 2 θ–cot 2 θ)+(sec 2 θ–tan 2 θ)+2+2 = k 1+1+1+2+2 = k k = 7 OR On putting θ = 45°, (sin 45° + cosec 45°) 2 + (cos 45° + sec 45°) 2 = k + tan 2 45° + cot 2 45° 2 2 2 2 1 1 2 2 k 1 1 2 2     + + + = + +         9 9 k 2 2 2 + = + , k = 7 531. Let 2sinx a 1 sin x cos x = + + + + + + + + and c b 1 sin x = + .Then a = b, if c = ? (a) 1 sin x cos x + − (b) 1 sin x cos x − (c) 1 sin x cos x + (d) 1 cos x sin x + − SSC CGL (TIER-I) – 06.06.2019 (Shift-II) Ans. (a) : Ùeefo x = 90 0 0 0 0 2sin90 a 1 sin90 cos 90 = + + 2 a 1 2 = = And, 0 c c c b 1 sin x 2 1 sin 90 = = = + + But a = b i.e. c 1 c 2 2 = ⇒ = From option (a), c = 1 + sinx – cosx c = 1+ sin90 0 – cos90 0 = 2 532. If 2 2 2 cos 3, cot cos θ = θ− θ θ− θ θ− θ θ− θ 0 0 < θ < 90 0 , then the value of cotθ + cosecθ is: (a) 3 2 (b) 3 (c) 2 3 − (d) 33 4 SSC CGL (TIER-I) – 06.06.2019 (Shift-II) Ans. (b) : 2 2 2 cos 3 cot cos θ = θ− θ ( ) 2 2 2 2 2 2 2 cos θ cos θ.sin θ =3 cot θ - cos θ cos θ 1-sin θ ⇒ 2 tan θ = 3 tanθ = 3 θ = 60º ⇒ ⇒ 0 0 cot cosec cot 60 cosec60 ∴ θ+ θ= + 1 2 3 3 3 3 3 = + = = 533. If 3sinθ = 2cos 2 θ, 0 0 < θ < 90 0 , then the value of (tan 2 θ + sec 2 θ – cosec 2 θ) is : (a) 7 3 (b) 7 3 − (c) –2 (d) 2 SSC CGL (TIER-I) – 10.06.2019 (Shift-I) Ans. (b) : 2 3sin 2 cos θ= θ Q (Given) ( ) { } 2 2 2 3sin 21 sin sin cos 1 θ= − θ θ+ θ= Q 2 3sin 2 2sin θ= − θ 2 2sin 3sin 2 0 θ+ θ− = 2 2sin 4sin sin 2 0 θ+ θ− θ− = ( ) ( ) 2sin sin 2 1 sin 2 0 θ θ+ − θ+ = ( )( ) sin 2 2sin 1 0 θ+ θ− = sin 2 0 or 2sin 1 0 θ+ = θ− = ( ) sin 2 Unconsidered θ=− 1 sin 2 θ= 0 sin sin 30 θ= 0 30 θ= 2 2 2 2 0 2 0 2 0 tan sec cosec tan 30 sec 30 cosec 30 ∴ θ+ θ− θ= + − 1 4 4 3 3 = + − 5 12 7 3 3 − − = =

Trigonometry 198 . 534. If x = 4cosA + 5sinA and y = 4sinA – 5cosA, then the value of x 2 + y 2 is: (a) 16 (b) 25 (c) 0 (d) 41 SSC CGL (Tier-I)– 03/03/2020 (Shift-I) Ans. (d) : x = 4 cosA + 5sinA ........ (i) y = 4 sin A – 5 cosA ....... (ii) By adding the square of equation (i) and (ii) x 2 +y 2 =16cos 2 A+25sin 2 A+40sinA.cosA+16sin 2 A+25 cos 2 A – 40sinA.cosA = 16(cos 2 A+sin 2 A) + 25 (sin 2 A+cos 2 A) = 16 + 25 = 41 535. If xcosA – ysinA = 1 and xsinA + ycosA = 4, then the value of 17x 2 + 17y 2 is (a) 289 (b) 49 (c) 7 (d) 0 SSC CGL (Tier-I) – 06/03/2020 (Shift-I) Ans. (a) : x cosA – ysinA = 1 ...........(i) x sinA + ycosA = 4 ..........(ii) By adding the square of equation (i) and (ii) x 2 cos 2 A + y 2 sin 2 A – 2yxsinA.cosA + x 2 sin 2 A + y 2 cos 2 A + 2yxsinA.cosA = 17 x 2 (cos 2 A + sin 2 A) + y 2 (sin 2 A + cos 2 A) = 17 x 2 + y 2 = 17 ∴ 17x 2 + 17y 2 = 289 536. What is the value of [(cos 3 2θ + 3 cos 2θ) ÷ (cos 6 θ – sin 6 θ)] ? (a) 0 (b) 1 (c) 4 (d) 2 SSC CGL (Tier-II) 18-02-2018 Ans. (c) : Given– 3 6 6 cos 2 3cos2 cos sin θ+ θ θ− θ Put θ = 30º 3 0 0 6 0 6 0 cos 60 3 cos 60 cos 30 sin 30 + + 1 1 1 1 3 2 2 2 2 3 3 3 1 1 1 4 4 4 4 4 4 × × + × = × × − × × 1 3 8 2 27 1 64 64 + = − 13 13 64 8 4 26 8 26 64 = = × = 537. If sin x = 1/2 and sin y = 2/3, then what is the value of [(6cos 2 x–4 cos 4 x)/(18 cos 2 y–27cos 4 y)]? (a) 27/20 (b) 15/14 (c) 25/21 (d) 17/14 SSC CGL (Tier-II) 9-3-2018 Ans. (a) : 1 2 sin x , sin y 2 3 = = Q ∴ 3 5 cos x , cosy 2 3 = = 2 2 2 4 2 4 2 2 2cos x 3 2cos x 6cos x 4cos x 18cos y 27cos y 9cos y 2 3cos y   − −   = −   −   3 3 3 3 2 3 2 27 4 4 2 2 1 5 5 20 5 9 2 3 3 9 9   × × − × ×     = = =   × × × − ×     538. What is the value of [(sin x + sin y) ( ) sin x sin y − ]/[(cosx + cosy) (cosy – cosx)]? (a) 0 (b) 1 (c) –1 (d) 2 SSC CGL (Tier-II) 17-2-2018 Ans. (b) : ( ) ( ) ( )( ) sin x sin y sin x sin y cos x cos y cos y cos x + − + − From value putting, By putting the value x = 90 0 and y = 0 0 ( )( ) ( )( ) 0 0 0 0 0 0 0 0 sin 90 sin0 sin 90 sin 0 cos90 cos0 cos0 cos90 + − = + − ( )( ) ( )( ) 1 0 1 0 1 1 0 1 1 0 1 + − = = = + − 539. If 2cos 2 θ + 3sinθ = 3, where 0 0 < θ < 90 0 , then what is the value of sin 2 2θ + cos 2 θ+ tan 2 2θ+ cosec 2 2θ ? (a) 29 6 (b) 29 3 (c) 35 6 (d) 35 12 SSC CGL (Tier-II) 13-09-2019 Ans. (c) : 2cos 2 θ + 3sinθ = 3 2(1–sin 2 θ) + 3sinθ–3 = 0 –2sin 2 θ + 3sinθ–1 = 0 2 sin 2 θ – 3sinθ + 1 = 0 2sin 2 θ – 2sinθ – sinθ + 1 = 0 2sinθ (sinθ–1) –1 (sinθ–1) = 0 (sinθ–1) (2sinθ–1) = 0 ∴ sinθ =1 θ = 90 0 (Invalid) [Q 0 0 < θ < 90 0 ] ∴ 1 sin 2 θ= θ = 30 0 sin 2 2θ + cos 2 θ+tan 2 2θ+cosec 2 2θ = sin 2 60 0 + cos 2 30 0 + tan 2 60 0 + cosec 2 60 0 3 3 4 3 4 4 3 = + + +

Trigonometry 199 . 3 13 2 3 = + 35 6 = 540. If θ lies in the first quadrant and cos 2 θ – sin 2 θ = 1 2 then the value of tan 2 2θ + sin 2 3θ is : (a) 4 3 (b) 3 (c) 7 2 (d) 4 SSC CGL (Tier-II) 12-09-2019 Ans. (d) : cos 2 θ – sin 2 θ = 1 2 cos2 θ = cos 60 0 2θ = 60 0 θ = 30 0 ∴ tan 2 2θ + sin 2 3θ = ( ) ( ) 2 0 2 tan 2 30 sin 3 30 0 × + × 2 0 2 0 tan 60 sin 90 = + ( ) () 2 2 3 1 = + = 3 + 1 = 4 541. ( ) ( ) – 2 2 2 2 1 cos sin ? cos ec 1 sin + θ + θ + θ + θ + θ + θ + θ + θ = θ θ θ θ θ θ θ θ (a) cosθ (1+sinθ) (b) secθ(1+sinθ) (c) 2cosθ(1+secθ) (d) 2secθ(1+secθ) SSC CGL (Tier-II) 11-9-2019 Ans. (d) : ( ) ( ) 2 2 2 2 1 + cos sin cosec - 1 sin θ + θ θ θ 2 2 2 2 2 1 cos 2cos sin cosec sin sin . + θ+ θ+ θ = θ θ− θ 2 1 2cos +1 1 - sin + θ = θ ( ) 2 21 cosθ cos θ + = 2 2 cosθ 1 2 cos θ cos θ   = +     ( ) 2 2 sec θ secθ = + ( ) 2secθ 1 secθ = + 542. If 7 sin 2 θ + 3cos 2 θ = 4, 0 0 < θ < 90 0 , then the value of (tan 2 2θ + cosec 2 2θ) is: (a) 7 (b) 15/4 (c) 13/3 (d) 13/4 SSC CPO-SI – 11/12/2019 (Shift-II) Ans. (c) 7 sin 2 θ + 3 cos 2 θ = 4 3 sin 2 θ +3cos 2 θ + 4sin 2 θ = 4 3(sin 2 θ + cos 2 θ) + 4sin 2 θ = 4 (3×1) + 4sin 2 θ = 4 4sin 2 θ = 1 sinθ = 1/2 θ = 30 0 ∴ tan 2 2θ + cosec 2 2θ = tan 2 60 0 + cosec 2 60 0 = ( ) 2 2 2 3 3   +     = 4 13 3 3 3 + = 543. If 3 + cos 2 θ = 3(cot 2 θ + sin 2 θ), 0º < θ < 90º, then what is the value of (cosθ + 2sinθ)? (a) 3 2 2 + (b) 3 2 (c) 2 3 1 2 + (d) 33 1 2 + SSC CPO-SI – 11/12/2019 (Shift-I) Ans. (c) 3 + cos 2 θ = 3(cot 2 θ + sin 2 θ) 3 + 1 – sin 2 θ = 3 [cosec 2 θ – 1 + sin 2 θ] 4 – sin 2 θ = 2 2 3 3 3sin sin − + θ θ 7 = 2 2 3 4sin sin + θ θ 4sin 4 θ – 4sin 2 θ – 3sin 2 θ + 3 = 0 4sin 2 θ (sin 2 θ – 1) – 3(sin 2 θ – 1) = 0 (sin 2 θ – 1) (4sin 2 θ – 3) = 0 ∴ sin 2 θ = 1 θ = 90º (Unconsidrable) [0 0 < θ < 90 0 ] ∴ sin 2 θ = 3 4 sinθ = 3 2 θ = 60º Hence cos60º + 2sin60º = 1 3 2 2 2 + × = 2 3 1 2 + 544. If secθ = 4x and ( ) 4 tan ,x 0 x θ= ≠ θ= ≠ θ= ≠ θ= ≠ then value of 2 2 1 8 x x         −                 is : (a) 1 8 (b) 1 2 (c) 1 16 (d) 1 4 SSC CHSL 05/07/2019 (Shift-II)

Trigonometry 200 . Ans. (b) : sec 4x θ= and 4 tan x θ= And ( ) 2 2 sec 4x θ= 2 2 sec 16x ...........(i) θ= and 2 2 16 tan ...........(ii) x θ= By subtracting equation (ii), from equation (i), 2 2 2 2 16 sec tan 16x x θ− θ= − { } 2 2 2 2 1 1 16 x sec tan 1 x   = − θ− θ=     Q 2 2 1 1 8x 2 x   = −     On dividing by 2 in both equations. Hence 2 2 1 1 8x 2 x   − =     545. If cos 2 θ – sin 2 θ – 3cosθ + 2 = 0, 0 0 < θ < 90 0 then the value of 4cosecθ + cotθ is : (a) 3 (b) 33 (c) 4 (d) 4 3 SSC CHSL 02/07/2019 (Shift-I) Ans. (b) 2 2 cos sin 3cos 2 0 θ− θ− θ+ = ( ) 2 2 cos 1 cos 3cos 2 0 θ− − θ− θ+ = 2 2 cos 1 cos 3cos 2 0 ⇒ θ− + θ− θ+ = 2 2 cos 3cos 1 0 ⇒ θ− θ+ = 2 2 cos 2 cos cos 1 0 ⇒ θ− θ− θ+ = ( ) ( ) 2 cos cos 1 1 cos 1 0 ⇒ θ θ− − θ− = ( )( ) 2 cos 1 cos 1 0 ⇒ θ− θ− = 1 cos , cos 1 2 θ= θ= 0 0 60 , 0 θ= θ= (Unconsidered) [Q0 0 <θ<90 0 ] ∴ 0 0 4cosec60 cot 60 + 2 1 9 3 4 3 3 3 3 = × + = × 33 = 546. If       1 1 - 1 + cosecθ 1 - cosecθ o o cosθ = 2, 0 < θ < 90 then the value sin 2 θ + cot 2 θ + sec 2 θ is : (a) 1 2 2 (b) 1 3 2 (c) 2 (d) 1 SSC CHSL 03/07/2019 (Shift-III) Ans. (b) : 1 1 cos 2 1 cos ec 1 cos ec   − θ=   + θ − θ   ( ) ( ) ( )( ) 1 cos ec 1 cos ec cos 2 1 cos ec 1 cos ec   − θ− + θ ⇒ θ=     + θ − θ   2 1 cos ec 1 cos ec cos 2 1 cos ec − θ− − θ   ⇒ θ=   − θ   2 2 cos ec cos 2 cot − θ   ⇒ θ=   − θ   2 2 2 sin cos 2 sin cos   θ ⇒ × θ=   θ θ   ⇒ 2 tan θ = 2 ⇒ tanθ = 1 ⇒ θ = 45 o ∴ sin 2 θ + cot 2 θ +sec 2 θ = sin 2 45 o +cot 2 45 o +sec 2 45 o () ( ) 2 2 2 1 1 2 2   = + +     1 1 2 2 = + + 1 3 2 = 547. If 2sin 2 θ + 5 cosθ - 4 = 0, 0 o < θ < 90 o then the value of tanθ + sinθ is : (a) 3 2 (b) 2 3 (c) 33 2 (d) 3 3 SSC CHSL 04/07/2019 (Shift-III) Ans. (c) : 2sin 2 θ + 5cosθ – 4 = 0, 0 o < θ < 90 o 2(1 – cos 2 θ) + 5cosθ – 4 = 0 2 – 2cos 2 θ + 5cosθ – 4 = 0 2cos 2 θ – 5cosθ + 2 = 0 cosθ = ( ) ( ) 2 5 5 4 2 2 2 2 −− ± − − × × × cosθ = 3 4 5− (On taking negative sign) cosθ = o 1 cos 60 2 = θ = 60 o ∴ tanθ + sinθ = tan60 o + sin60 o 3 3 2 = + 33 2 =

Trigonometry 201 . 548. for 0 0 < θ < 90 0 , If 2cos 2 θ = 3sinθ then the value of (cosec 2 θ – cot 2 θ + cos 2 θ) is : (a) 1 1 2 (b) 3 1 4 (c) 1 2 4 (d) 3 2 4 SSC CHSL 04/07/2019 (Shift-I) Ans. (b) : 0 0 < θ < 90 0 Given, 2 2 cos 3sin θ= θ ( ) 2 21 sin 3sin − θ= θ 2 2sin 3sin 2 0 θ+ θ− = ( )( ) 2sin 1 sin 2 0 θ− θ+ = 1 sin sin 2 2 θ= θ=− (Unconsidered) 0 30 θ= Hence cosec 2 θ – cot 2 θ + cos 2 θ = cosec 2 (30 0 ) – cot 2 (30 0 ) + cos 2 30 0 ( ) ( ) 2 2 2 3 2 3 2   = − +       3 4 3 4 = − + 3 3 1 1 4 4 = + = 549. If secθ = 3x and ( ) 3 tan θ ,x 0 x = ≠ = ≠ = ≠ = ≠ then the value of 2 2 1 9 x x         −                 is : (a) 1 2 (b) 1 (c) 1 3 (d) 1 4 SSC CHSL 05/07/2019 (Shift-III) Ans. (b) : 2 2 secθ 3x sec θ 9x = ⇒ = (By squaring the both side) 2 2 3 9 tan θ tan θ x x = ⇒ = Q 2 2 sec θ tan θ 1 − = 2 2 9 9x 1 x ⇒ − = 2 2 1 9x 1 x   ⇒ − =     550. If A = 2(sin 6 θ + Cos 6 θ) – 3 (sin 4 θ + cos 4 θ) then the value of 3α such that cosα = 3+A 5+A is: (a) 135 o (b) 45 o (c) 180 o (d) 90 o SSC CHSL –13/10/2020 (Shift-I) Ans. (a) : ( ) ( ) 6 6 4 4 A 2 2sin cos 3 sin cos = θ+ θ− θ+ θ ( ) ( ) 3 2 2 3 4 4 A 2 sin (cos ) 3 sin cos   = θ + θ − θ+ θ     ( )( ) 2 2 4 4 2 2 A 2 sin cos sin cos sin cos   = θ+ θ θ+ θ− θ θ   ( ) 4 4 3 sin cos − θ+ θ 4 4 2 2 4 4 A 2sin 2cos 2sin .cos 3sin 3cos = θ+ θ− θ θ− θ− θ ( ) 2 2 2 A sin cos =− θ+ θ A = –1 3 A 1 cos 5 A 2 + α= = + cos cos 45 α= ° α = 45° 3α = 3 × 45° = 135° 551. If a sinA + b cosA = c, then a cosA – b sinA is equal to: (a) 2 2 2 a b c − − (b) 2 2 2 a b c − + (c) 2 2 2 a b c + − (d) 2 2 2 a b c + + SSC CHSL –18/03/2020 (Shift-I) Ans. (c) : a sinA + b cosA = c ____ (i) Let a cosA – b sinA = x ____(ii) By squaring the equation (i), 2 2 2 2 2 a sin A b cos A 2ab sin A cos A c + + = ___(iii) By squaring the equation (ii) 2 2 2 2 2 a cos A b sin A 2ab sin A. cos A x + − = ____(iv) By adding the equation (iii) and (iv) ( ) ( ) 2 2 2 2 2 2 2 2 a sin A cos A b sin A cos A c x + + + = + 2 2 2 2 c x a b + = + 2 2 2 2 x a b c = + − x = 2 2 2 a b c + − 552. If sinθ + cosecθ = 2 then find the value of (sin 153 θ +cosec 253 θ) is : (a) 253 153 (b) 1 153 253 × (c) 153 253 (d) 2 SSC CHSL 10/07/2019 (Shift-I) Ans. (d): sin cosec 2 θ+ θ= 0 0 sin 90 cosec90 2 ∴ + = On putting the value of θ = 90º ( ) 1 1 2 + = 2 = 2 L.H.S = R.H.S (is true) ( ) ( ) 153 253 153 253 0 0 sin cosec sin 90 cosec 90 ∴ θ+ θ= + = 1 + 1 = 2

Trigonometry 202 . 553. If 3 cos x 2 − = and 3 x 2 π π< < π< < π< < π< < then find the value of 2 2 2cot x 3cosec x + : (a) 8 (b) 16 (c) 14 (d) 18 SSC CHSL 09/07/2019 (Shift-I) Ans. (d) : 3 cos x 2 − = 3 x 2 π π< < or180 x 270 °< < o 0 cos x cos 30 =− ( ) 0 0 cos x cos 180 30 ⇒ = + 0 cos x cos 210 ⇒ = 0 x 210 ⇒ = 2 2 2 0 2 0 2cot x 3cosec x 2 cot 210 3cosec 210 + = × + ( ) ( ) 2 0 0 2 0 0 2 cot 180 30 3cosec 180 30 = + + + ( ) ( ) 2 2 2 3 3 2 = × + ×− 6 12 18 = + = 554. If cot θ = 5x and 5 cosec (x 0) x θ= ≠ θ= ≠ θ= ≠ θ= ≠ then value of 2 2 1 5 x x         −                 : (a) 1 5 (b) 1 2 (c) 1 4 − (d) 1 5 − SSC CHSL08/07/2019 (Shift-III) Ans. (d) : Given– cot θ = 5x and 5 cosec x θ= ⇒ 2 2 cot 25x θ= .............(i) And 2 2 25 cosec x θ= ............(ii) From equation (i) and (ii), 2 2 2 2 25 cot cosec 25x x θ− θ= − ⇒ 2 2 2 2 1 cot cosec 25 x x   θ− θ= −     ⇒ 2 2 1 1 5 5x x   −= × −     {1+cot 2 θ = cosec 2 θ} ⇒ 2 2 1 1 5x 5 x   − =−     555. If θ+ 5cot 3 cosec θ 2 3 cosec θ + 3cot θ = 1, 0° < θ < 90°, then the value of θ− 2 2 2 2 7 3 cot cosec θ 2 4 3 4 sin θ+ tan θ 2 will be: (a) 2 (b) 7 (c) 5 (d) 3 SSC CHSL 10/082021 (Shift-II) Ans. (c) : 5cot 3 cosec 2 3 cosec 3cot θ+ θ θ+ θ = 1 5cot 3 cosec 2 3 cosec 3cot θ+ θ= θ+ θ 2 cot 3 cosec θ= θ 3 cos cos 30º 2 θ= = 30º θ= 2 2 2 2 2 2 2 2 7 3 7 3 cot cosec cot 30º cos ec 30º 2 4 2 4 3 3 4sin tan 4sin 30º tan 30º 4 2 θ− θ × − = θ+ θ + 7 3 15 3 4 2 4 2 5 1 3 1 3 4 4 2 3 2 × − × = = = × + × 556. If cot 2 θ+cot 4 θ = 2, then the value of 2sin 4 θ + sin 2 θ is: (a) 1 (b) 3 (c) 5 (d) 2 SSC CHSL 11/082021 (Shift-II) Ans. (a) : cot 2 θ + cot 4 θ= 2 On putting the value of θ = 45º L.H.S = cot 2 45º + cot 4 45º= 1 +1 = 2 = R.H.S So, 2sin 4 θ + sin 2 θ = 2×sin 4 45º+sin 2 45º 1 1 1 1 2 1 4 2 2 2 = × + = + = 557. If 5k = tan θ and 5 k = secθ, then what is the value of 10 2 2 1 k k   −     ? (a) 2 5 (b) 2 5 − (c) –2 (d) 2 SSC CHSL 16/04/2021 (Shift-III) Ans.(b) : tan 5k tan k (i) 5 θ = θ⇒ = → 5 1 sec sec (ii) k k 5 θ = θ⇒ = → From equation (i) and (ii) 2 2 2 2 1 tan sec 1 k k 25 25 25 θ θ − − = − = ( ) 2 2 sec tan 1 θ− θ= Q 2 2 1 1 2 10 k 10 k 25 5 −   ∴ − = × =     558. What is the value of             π 3π tan + A × tan +A 4 4 ? (a) 1 (b) 0 (c) cot A/2 (d) –1 SSC CGL (Tier-II) 18-02-2018

Trigonometry 203 . Ans. (d) : 3 tan A tan A 4 4 π π     + × +         Q tan(A+B) = tan A tan B 1 tan A. tan B + − ∴ 3 tan tan A tan tan A 4 4 3 1 tan . tan A 1 tan tan A 4 4 π π + + × π π − − 3 tan 1, tan 1 4 4 π π =− = Q = 1 tan A 1 tan A 1 tan A 1 tan A + −+     ×     − +     = –1 559. Which is increased continuously in the range 0º < θ > 90º (a) cosecθ (b) cosθ (c) cotθ (d) tanθ SSC CHSL (Tier-I) 09/07/2019 (Shift-III) Ans. (d) : The value of sinθ is increased between 0º to 90º and the value of cosθ is decrease b/w 0º to 90º Hence the value of cosecθ is decrease and the value of secθ is increase continuously. sin tan cos θ θ= θ Q Hence, the value of tanθ will be increased and the value of cot will be decreased continuously. 560. For all ∝ i (i = 1, 2, 3, .....20) between 0 0 and 90 0 given that cos∝ 1 + cos∝ 2 + cos∝ 3 + ........+ cos∝ 20 = 20 then find the value of (∝ 1 + ∝ 2 + ∝ 3 + .......+ ∝ 20 ) : (a) 0 0 (b) 900 0 (c) 1800 0 (d) 20 0 SSC CHSL (Tier-I) 11/07/2019 (Shift-II) Ans. (a) : Maximum value of cos∝ = 1 ∝ = 0 0 cos∝ 1 + cos∝ 2 + ........+ cos∝ 20 = 20 is possible when, cos∝ 1 = cos∝ 2 = ...... = cos∝ 20 = 1 ∝ 1 = ∝ 2 = ...... = ∝ 20 = 0 0 ∴ ∝ 1 + ∝ 2 + ....... + ∝ 20 = 0 + 0 + ...... + 0 = 0 0 561. Between 0 0 and 90 0 for all α i (i = 1, 2, 3, 4, .....20) given that sinα 1 + sinα 2 + sinα 3 + ..... + sinα 20 = 20 then what is the value of (α 1 + α 2 + α 3 + ......+ α 20 ) in degree. (a) 20 (b) 1800 (c) 900 (d) 0 SSC CHSL (Tier-I) 11/07/2019 (Shift-I) Ans. (b) : we know that the maximum value of sinθ=1 1 2 3 20 sin sin sin ...... sin 20 α+ α+ α+ + α = , By taking the value of 0 1 2 3 20 ...... 90 α =α =α α = 0 0 0 0 sin 90 sin 90 sin 90 ........ sin 90 20 + + + = ⇒ 1 + 1 + 1 + ...........+ 1 = 20 20 20 ⇒ = 1 2 3 20 ...... 90 20 1800 ∴ α +α +α + +α = × = 562. If x = cosecA + cosA and y = cosecA – cosA, then find the value of             2 2 2 x-y + -1 x+y 2 . (a) 2 (b) 1 (c) 0 (d) 3 SSC CHSL –19/10/2020 (Shift-I) Ans. (c) : x = cosec A + cos A, y = cosec A – cos A x + y = 2 cosec A x – y = 2 cosA ∴ 2 2 2 2 2 x y 2 2 cos A 1 1 x y 2 2cosec A 2     −     + − = + −         +         ( ) 2 2 sin A cos A 1 = + − = 1 – 1 = 0 563. The general solution of the equation       π tan3x + cot 2x + =0 3 is: (a) 2n ,n z 3 π π± ∈ (b) 2n ,n z 6 π π± ∈ (c) n ,n z 3 π π+ ∈ (d) 5 n ,n z 6 π π+ ∈ SSC Sel. Post Phase VIII (G.L.) 09.11.20 (Shift-2) Ans. (d) : tan 3x cot 2x 0 3 π   + + =     tan 3x cot 2x 3 π   =− +     tan 3x cot 2x 3 π   = − −     tan 3x tan 2x 2 3 π π    = − − −         tan 3x tan 2x 2 3 π π   = + +     or 5 3x 2x 6 π = + 5 x 6 π = So its general solution is 5 x n 6 π = π+ ,n z ∈ Powered by TCPDF (www.tcpdf.org)

Height and Distance 204 . 03. Height and Distance Based On TCS Pattern Typewise Exam Question No. Years Type-I Height and Distance CGL (Tier-1) 25 (2017–2023) CGL (Tier-2) 7 CHSL (Tier-1) 30 CHSL (Tier-2) 6 Selection Post VII, VIII, XI 3 SSC MTS 6 SSC GD – SSC CPO 13 Trend Analysis of Questions topicwise from CGL (Pre & Mains) CHSL (Pre & Mains) Selection Post VII, VIII, XI, SSC MTS, SSC GD & Other Exams (2017-2023)