JEE Main2017MathematicsInverse Trigonometric FunctionsActual
The value of tan - 1 ⁡ 1 + x 2 + 1 - x 2 1 + x 2 - 1 - x 2 , x < 1 2 , x ≠ 0 , is equal to:
Options
- Aπ 4 + 1 2 cos - 1 x 2
- Bπ 4 - cos - 1 x 2
- Cπ 4 - 1 2 cos - 1 x 2
- Dπ 4 + cos - 1 x 2
Correct answer
A. π 4 + 1 2 cos - 1 x 2
Step-by-step solution
Assume that , x 2 = cos ⁡ 2 θ ; θ = 1 2 cos -1 x 2 ∴ tan - 1 ⁡ 1 + cos ⁡ 2 θ +   1 - cos ⁡ 2 θ 1 + cos ⁡ 2 θ -   1 - cos ⁡ 2 θ = tan - 1 ⁡ 2 cos 2 θ +   2 sin 2 θ 2 cos 2 θ -   2 sin 2 θ = tan - 1 ⁡ cos θ +   sin θ cos θ -   sin θ = tan - 1 ⁡ 1 + tan ⁡ θ 1 - tan ⁡ θ = tan - 1 ⁡ tan ⁡ π 4 + θ = π 4 + 1 2 cos -