Question

 If 2sin(3x -15)° = 1, 0° < (3x - 15) < 90°, then find the value of cos²(2x +15)° + cot² (x + 15)°.
1. \(\frac{7}{2}\)
2. \(\frac{5}{2}\)
3. 1
4. \(-\frac{7}{2}\)

Answer 1

Correct Answer - Option 1 : \(\frac{7}{2}\)

Given:

⇒ 2sin(3x -15)° = 1

Calculation:

⇒ sin(3x - 15)° = 1/2

⇒ (3x  -15)° = sin^-1(1/2)

But, 0° < (3x - 15) < 90° then,

⇒ (3x - 15)° = 30°

⇒ x = 15°

Then,

⇒ ? = cos²(2x +15)° + cot2 (x +15)°

⇒ ? = cos²(45)° + cot²(30)°

⇒ ? = 1/2 + 3

⇒ ? = 7/2

∴ cos2(2x +15)° + cot2 (x +15)° = 7/2