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°
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