Trigonometry (Railway Maths) (Part – III)

Total Questions: 50

1. If 2 sin y + cos y = √5 sin y, then find the value of tan y. [Level 6 (09/05/2022) Shift 1]

Correct Answer: (c) √5 + 2
Solution:

2. Evaluate [Level 5 (12/06/2022) Shift 1]

Correct Answer: (a) −1/2
Solution:

3. If cot²θ = 1 + cos²θ − sin²θ, 0° < θ < 90°, then find the value of tan²θ + cosec²θ. [Level 5 (12/06/2022) Shift 2]

Correct Answer: (d) 3
Solution:

4. Find the answer of given question? [Level 2 (13/06/2022) Shift 1 ]

Correct Answer: (b) 30°
Solution:

5. If 3 cos²θ + 1 = 4 sinθ, 0° < θ < 90°, then the value of sec²θ is: [Level 3 (14/06/2022) Shift 2]

Correct Answer: (a) 9/5
Solution:

3cos²θ + 1 = 4sinθ
3(1 − sin²θ) + 1 = 4 sinθ
3 − 3 sin²θ + 1 = 4 sinθ
3 sin²θ + 4 sinθ − 4 = 0
3 sin²θ + 6 sinθ − 2 sinθ − 4 = 0
3 sinθ (sinθ + 2) − 2 (sinθ + 2) = 0
(3 sinθ − 2) = 0
⇒ sinθ = 2/3 = P/H
B = √(3² − 2²) = √5
sec²θ = (H/B)² = (3/√5)² = 9/5

6. If sin²β − sin 30° = 0 and β is an acute angle, find the value of β. [Level 5 (15/06/2022) Shift 3]

Correct Answer: (d) 45°
Solution:

sin²β − sin 30° = 0
By options, when
β = 45°; sin²45° − sin 30°
= 1/2 − 1/2 = 0

7. If tanθ + 3 cotθ = 2√3, 0° < θ < 90°, then what is the value of cosec²θ? [Level 3 (17/06/2022) Shift 1]

Correct Answer: (c) 4/3
Solution:

tanθ + 3cotθ = 2√3
1/cotθ + 3cotθ = 2√3
1 + 3cot²θ = 2√3 cotθ
3cot²θ − 2√3 cotθ + 1 = 0
Then, cotθ = 1/√3
cosec²θ = 1 + cot²θ = 1 + (1/√3)²
= 1 + 1/3 = 4/3

8. If sin = 3/4 and cosθ = 5/4, then the value of 1 + tanθ / 1 - cotθ is: [RRB NTPC 28/12/2020 (Morning) ]

Correct Answer: (d) -12/5
Solution:

9. Find the answer of given question? [RRB NTPC 28/12/2020 (Morning) ]

Correct Answer: (a) 60°
Solution:

10. If cos2θ = sinθ and θ lies between 0 and 90°, then θ will be: [RRB NTPC 28/12/2020 (Evening) ]

Correct Answer: (c) 30°
Solution:

As cos2θ = sinθ and θ lies between 0° and 90°.
⇒ cos2θ = cos(90° − θ)
⇒ 3θ = 90° ⇒ θ = 30°
Then θ will be 30°.