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SSC CGL TIER-1 Previous Year Trigonometry TCS Questions & Solutions

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SSC CGL TIER-1 Previous Year Trigonometry TCS Questions & Solutions 

SSC CGL T-2 2019 SSC CGL T-1 2019 SSC CHSL 2019 SSC CPO 2020 SSC MTS 2019
SSC CGL T-2 2018 SSC CGL T-1 2018 SSC CHSL 2018 SSC CPO 2019 SSC MTS 2017

Q1. If x = 4cosA + 5sinA and y = 4 sinA - 5 cosA,then the value of

x2 +y2 is

यदि x = 4cosA + 5sinA तथा y = 4 sinA - 5 cosA है, 

तो x2 +y2 का मान क्या होगा ? 

SSC CGL 3 March 2020 (Morning)

(a) 41

(b) 0

(c) 16

(d) 25

Q2. If A lies in the first quadrant and 6tanA = 5, then the value

is

यदि A प्रथम चतुर्थांश में स्थित है तथा 6tanA = 5 है, तो  का मान क्या होगा ? 

SSC CGL 3 March 2020 (Morning)

(a) -2

(b) 1

(c) 16

(d) 4

Q3. If A+B = 45°, then the value of 2(1+ tanA)(1+ tanB) is:

यदि A+B = 45° है, तो 2(1+ tanA)(1+ tanB) का मान होगा : 

SSC CGL 3 March 2020 (Morning)

(a) 2

(b) 4

(c) 1

(d) 0

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Q4. If  2sinθ+15 cos2 θ=7, 0°<θ<90°, than tanθ+cosθ+ secθ =

यदि  2sinθ+15 cos2 θ=7, 0°<θ<90° है, तो tanθ+cosθ+ secθ = ?

SSC CGL 3 March 2020 (Afternoon)

  (d) 4

Q6. The value of the expression cosec(85°+θ) - sec(5°-θ) - tan(55°+θ) + cot(35°- θ) is:

व्यंजक cosec(85° + θ) - sec(5°-θ) - tan(55°+θ) + cot(35°- θ) का मान है : 

SSC CGL 3 March 2020 (Afternoon)

(a)3/2 

(b) 0

(c) -1

(d)1

Q8. If 2 sinθ - 8 cos2θ + 5 = 0, 0°<θ<90°, then what is the value of (tan 2θ + cosec 2θ)?

यदि 2 sinθ - 8 cos2θ + 5 = 0, 0°<θ<90° है, तो (tan 2θ + cosec 2θ) का मान क्या होगा ? 

SSC CGL 3 March 2020 (Evening)

Q9. In the figure, what is the value of cotθ?

इस आकृति में, cotθ का मान कितना है? 

Q10. If secθ - tanθ =,(0<x<y) and 0°<θ<90°, then sinθ is equal to:

यदि secθ - tanθ =   (0<x<y) तथा 0°<θ<90° है, तो sinθ का मान किसके बराबर है ? 

SSC CGL 4 March 2020 (Morning)

SSC CGL 4 March 2020 (Afternoon)

(a) -2

(b) 1

(c) 2

(d) -1

Q16. If secθ + tanθ = p, 0°<θ<90°, then

is equal to:

यदि secθ + tanθ = p है, जहाँ  0°<θ<90° है, तो

  का मान किसके बराबर होगा ? 

SSC CGL 4 March 2020 (Evening)

(a) cosecθ

(b) sinθ

(c) cosθ

(d) 2cosecθ

 

SSC CGL 5 March 2020 (Afternoon)

(a) 0

(b) 2

(c) -1

(d) 1

Q26. The value of cos 0°cos 30°cos 45°cos 60°cos 90° is:

cos 0°cos 30°cos 45°cos 60°cos 90° का मान क्या होगा? 

SSC CGL 5 March 2020 (Evening)

(a) 5

(b) 

(c) 3

(d) 0

Q28. If xcosA - ysinA = 1 and xsinA + ycosA = 4, then the value of 17x2 +17y2 is:

यदि xcosA - ysinA = 1 तथा xsinA + ycosA = 4 है, तो  17x2 +17y2 का मान क्या होगा ? 

SSC CGL 6 March 2020 (Morning)

(a) 0

(b) 49

(c) 7

(d) 289

Q29. If (2sin A + cosec A) = 2√2, 0°<A<90°, then the value of 2(sin4A+cos4 A) is:

यदि (2sin A + cosec A) = 2√2, 0°<A<90° है, तो  2(sin4A+cos4 A) का मान क्या होगा ? 

SSC CGL 6 March 2020 (Morning)

(a) 1

(b) 2

(c) 4

(d) 0

SSC CGL 6 March 2020 (Morning)

(a) -1

(b) -2

(c) 1

(d) 2

Q31. Solve the following/ हल कीजिए : 

 sin 0° sin 30° sin 45° sin 60° sin 90° = ?

SSC CGL 6 March 2020 (Afternoon)

(a) 0

(b) 

(c) 1

(d) 4

Q32. If (cos2θ-1)(1+tan2 θ) +2tan2 θ  = 1, 0°<θ<90°, then θ is 

यदि  (cos2θ-1)(1+tan2 θ) +2tan2 θ  = 1 है, जहाँ 0°<θ<90°, तो  θ का मान होगा : 

SSC CGL 6 March 2020 (Afternoon)

(a) 45°

(b) 60°

(c) 30°

(d) 90°

 

SSC CGL 6 March 2020 (Evening)

(a) 2

(b) 4

(c) 8

(d) 1

Q37. The value of is:

का मान है : 

SSC CGL 7 March 2020 (Morning)

(a) -2sin2θcos2θ

(b) -1

(c) 0

(d) 1

Q38. What is the value of sin 30°+cos 30°-tan 45°?

sin 30°+cos 30°-tan 45° का मान क्या होगा?

SSC CGL 7 March 2020 (Morning)

Q39. In the given figure, cosθ is equal to: 

दी गयी आकृति में, cosθ  किसके बराबर है ? 

Q40. If 3sec2 θ+tanθ=7, 0°<θ<90°, then the value of 

Q42. If cotθ + tanθ = 2secθ, 0°<θ<90°, then the value of

Q43. If 5cos2 θ+1=3sin2θ, 0°<θ<90°, then what is the value of 

Q44. Solve the following/ निम्नलिखित को हल कीजिए  

SSC CGL 7 March 2020 (Evening)

(a) 2

(b) 1

(c) 0

(d) -1

Q45. In the right triangle shown in the figure, what is the value of cosecθ?

 इस आकृति में दिए गए समकोण त्रिभुज में, cosecθ का मान क्या है ? 

Q46. If 6tanθ-5√3secθ+12 cotθ = 0, 0°<θ<90°, then the value of (cosecθ+secθ) is: 

यदि 6tanθ-5√3secθ+12 cotθ = 0, 0°<θ<90° है, तो (cosecθ+secθ) का मान क्या होगा ? 

SSC CGL 9 March 2020 (Morning)

Q47. The value of (cosecA + cotA +1)(cosecA - cotA + 1) - 2cosecA is: 

(cosecA + cotA +1)(cosecA - cotA + 1) - 2cosecA का मान है : 

SSC CGL 9 March 2020 (Afternoon)

(a) 4cosecA

(b) 2

(c) 2cosecA

(d) 0

Q48. If 5 cotθ = 3, find the value of 

Q49. Solve the following/ हल कीजिए | 

SSC CGL 9 March 2020 (Afternoon)

(a) 1
(b) 3

(c) 0
(d) 2

Q50. The value of is: 

का मान क्या होगा ? 

SSC CGL 9 March 2020 (Evening)

(a) 4

(b) 3

(c) 1

(d) 2

Q51. The value of cos 10° cos 30° cos 50° cos 70° cos 90° is:

cos 10° cos 30° cos 50° cos 70° cos 90° का मान है :

SSC CGL 9 March 2020 (Evening)

(a) 3

(b) 0

(c) 5

(d) 1

Q52. The value of (cosec 30° - tan 45°)cot 60° tan 30° is:

(cosec 30° - tan 45°)cot 60° tan 30° का मान क्या है?  

SSC CGL 9 March 2020 (Evening)

(a) 13

(b) 1

(c) 3

Here are the solutions to the above questions of trigonometry asked in the SSC exam asked by TCS. Some questions have alternate solutions.

Sol 1. (a)  x= 4cosA + 5sinA

y= 4sinA - 5cosA

x2 = 16cos2 A + 25sin2 A + 40 cosA sinA

y2 = 16sin2 A + 25cos2 A - 40 cosA sinA

x2 +y2 = 16(cos2 A + sin2 A) + 25(sin2 A + cos2 A)= 41

Sol 2. (b) 

Sol 3. (b) We know that, if x+y = 45° than (1+ tanx)(1+ tany) = 2

Here A+B= 45°

Then,

2(1+ tanA)(1+ tanB)= 4

Sol 4. (a)  2sinθ+15 cos2θ=7, 0°<θ<90°

  2sinθ+15(1-sin2θ)=7

15 sin2θ-2sinθ-8=0

Sol 5. (b) It is given that

Sol 6. (b) Put θ= 5° in given equation: cosec(85°+θ)-sec(5°-θ)-tan(55°+θ)+cot(35°-θ)

We get:

Cosec 90°- sec0° - tan60° + cot30°

1-1-√3+√ 3

0

Sol 7. (a)

Sol 8. (d) It is given that 

2 sinθ - 8 cos2θ + 5 = 0

And 0°<θ<90°

cos2θ= 1- sin2θ

2 sinθ - 8 cos2θ + 5 = 0

2 sinθ - 8(1-sin2θ)+5 = 0

8sin2θ + 2 sinθ - 3 =0

Solve the quadratic equation, we get

Sinθ = 1/2 or -3/4

Sinθ can not be negative for 0°<θ<90°

Thus θ= 30°

Now, tan 2θ + cosec 2θ = tan 60°+ cosec 60°

Sol 14. (d) 3sec2θ tan2θ + tan6 θ- sec6θ = ?

We know that: sec2θ-tan2θ = 1

Cubing both sides we get:

Sec6θ - tan6θ - 3sec2θtan2θ = 1

Thus,

 3sec2θ tan2θ + tan6θ- sec6θ = -1

Only option (c) satisfies.

Sol 16. (b) secθ+tanθ = p

Sol 17. (a) 7cos2θ+3sin2θ=6, 0°<θ<90°

sin2θ = 1-cos2θ

4cos2θ = 3

Sol 19.

 (d) 12cos2θ-2sin2θ+3cosθ = 3, 0°<θ<90°

Put sinθ = 1 - cos2θ

We get:

14 cos2θ + 3 cosθ -5 = 0

Sol 22. (b) 11sin2θ-cos2θ+4sinθ-4=0 , 0°<θ<90°

Cos2θ = 1 - sin2θ

11 sin2θ -1 + sin2θ + 4 sinθ - 4 = 0

12 sin2θ + 4 sinθ - 5 = 0

Sinθ = 1/2θ = 30°

Sol 26. (d) cos 0°cos 30°cos 45°cos 60°cos 90° = 0

Sol 27. (d) tanθ-cotθ = cosecθ, 0°<θ<90°

At θ = 60°, tanθ-cotθ = cosecθ

[cos 90° = 0]

Sol 28. (d) xcosA - ysinA = 1 and xsinA + ycosA = 4

(x2+y2)(sin2A+cos2A) = (1+16) = 17

17(x2+y2) = 289

Sol 29. (a) 2sin A + cosec A = 2√2, at A = 45°

Then, 2(sin4 A+cos4A) = 1

Sol 31. (a) sin 0° sin 30° sin 45° sin 60° sin 90° = 0 (sin 0° = 0)

Sol 32. (a) (cos2θ-1)(1+tan2θ) +2 tan2θ = 1, 0°<θ<90°

(-sin2θ)(sec2θ) +2 tan2θ = 1

-tan2θ+2tan2θ = 1

tan2θ = 1

θ = 45°

Sol 33. (a)

Sol 35. (b) In the third quadrant tanA and cotA are positive.

Sol 38. (a) sin 30°+ cos 30°- tan 45°

Sol 47. (b) ([cosecA+cotA]+1)([cosecA-cotA]+1)-2cosecA

cosec2A-cot2A+cosecA+cotA+cosecA-cotA+1-2cosecA

1+2cosecA+1-2cosecA

2

Sol 51. (b) cos 90° = 0 

Therefore, 

cos10°cos30°cos50°cos70°cos90° = 0

Sol 52 (a) (cosec 30°-tan 45°)cot 60°tan 30° 

 SSC previous year questions 

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