Trigonometry (Railway Maths) (Part – I)

Total Questions: 50

41. 1 + tan 15° cot 75° is equal to: [NTPC CBT-I 20/01/2021 (Morning)]

Correct Answer: (a) sec²15°
Solution:

1 + tan 15° cot 75°
= 1 + tan² 15° = sec² 15°

42. If 3x = cosecθ and cotθ = 3/x, then the value of 9(x² − 1/x²) is: [NTPC CBT-I 23/01/2021 (Morning)]

Correct Answer: (a) 1
Solution:

43. The principal value of cot⁻¹(−1/√3) is: [NTPC CBT-I 29/01/2021 (Morning)]

Correct Answer: (b) 5π/6
Solution:

44. The minimum value of 4sin²θ + 5cos²θ is: [NTPC CBT-I 29/01/2021 (Evening)]

Correct Answer: (c) 4
Solution:

45. Angle 54° is equivalent to (in radians) : [NTPC CBT - I 04/02/2021 (Evening) ]

Correct Answer: (c) 3π/10
Solution:

54° = 54/180π = 3π/10

46. Simplify the following. [NTPC CBT - I 08/02/2021 (Evening) ]

Correct Answer: (c) 2cosθ
Solution:

47. Solve the following [NTPC CBT - I 11/02/2021 (Evening) ]

Correct Answer: (c) 112/65
Solution:

48. If cos(α + β) = 0, then sin(α − β) can be reduced to: [NTPC CBT-I 12/02/2021 (Morning)]

Correct Answer: (d) cos2β
Solution:

49. In triangle ABC, if sinA cosB = 1/4 and 3tanA = tanB then cot²A is equal to: [NTPC CBT-I 15/02/2021 (Evening)]

Correct Answer: (a) 3
Solution:

50. In a triangle ABC, tanA + tanB + tanC = ? [NTPC CBT - I 17/02/2021 (Morning) ]

Correct Answer: (c) tanAtanBtanC
Solution:

If A + B + C = 180°, then
tanA + tanB + tanC = tanAtanBtanC