NTA Abhyas JEE Main2020MathematicsInverse Trigonometric FunctionsPractice
The value of t a n - 1 1 - sin ⁡ x + 1 + sin ⁡ x 1 - sin ⁡ x - 1 + sin ⁡ x   ∀ x ∈ 0 , π 2 is equal to
Options
- Ax 2 - π 2
- Bx 2 + π 2
- Cx 2 - π
- Dπ 2 - x 2
Correct answer
A. x 2 - π 2
Step-by-step solution
t a n - 1 1 - sin ⁡ x + 1 + sin ⁡ x 1 - sin ⁡ x - 1 + sin ⁡ x = t a n - 1 1 - sin ⁡ x + 1 + sin ⁡ x 1 - sin ⁡ x - 1 + sin ⁡ x 1 - sin ⁡ x + 1 + sin ⁡ x 1 - sin ⁡ x + 1 + sin ⁡ x = t a n - 1 1 - sin ⁡ x + 1 + sin ⁡ x + 2 1 - s i n 2 x 1 - sin ⁡ x - 1 + sin ⁡ x = t a n - 1 2 1 + cos ⁡ x - 2 sin ⁡ x = t a n - 1 - 2 c o s 2 x 2 2 s i n x 2 c o s x 2 = t a n - 1 - c o t x 2 = t a n - 1 c o t π - x 2 = π