If , then the value of is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Let c = cos 2x. Then cos²x = (1+c)/2 and tan²x = sin²x/cos²x = (1-cos2x)/(1+cos2x) = (1-c)/(1+c).
Substitute: 5[(1-c)/(1+c) - (1+c)/2] = 2c + 9
Multiply through by 2(1+c): 5[2(1-c) - (1+c)²] = (2c+9)·2(1+c)
Left: 5[2-2c - 1-2c-c²] = 5[1-4c-c²]
Right: 2(2c+9)(1+c) = 2(2c+2c²+9+9c) = 2(2c²+11c+9)
So: 5 - 20c - 5c² = 4c² + 22c + 18
⟹ 9c² + 42c + 13 = 0 ⟹ (9c+3)(c+... ) — use quadratic: c = (-42 ± √(1764-468))/18 = (-42 ± 36)/18
So c = -6/18 = -1/3 or c = -78/18 (rejected since |cos2x| ≤ 1).
Thus cos2x = -1/3, and cos4x = 2cos²2x - 1 = 2(1/9) - 1 = -7/9.
Substitute: 5[(1-c)/(1+c) - (1+c)/2] = 2c + 9
Multiply through by 2(1+c): 5[2(1-c) - (1+c)²] = (2c+9)·2(1+c)
Left: 5[2-2c - 1-2c-c²] = 5[1-4c-c²]
Right: 2(2c+9)(1+c) = 2(2c+2c²+9+9c) = 2(2c²+11c+9)
So: 5 - 20c - 5c² = 4c² + 22c + 18
⟹ 9c² + 42c + 13 = 0 ⟹ (9c+3)(c+... ) — use quadratic: c = (-42 ± √(1764-468))/18 = (-42 ± 36)/18
So c = -6/18 = -1/3 or c = -78/18 (rejected since |cos2x| ≤ 1).
Thus cos2x = -1/3, and cos4x = 2cos²2x - 1 = 2(1/9) - 1 = -7/9.
[JEE Main 2017]