Let f x = sin tan - 1 x + sin cot - 1 x 2 - 1 , x > 1 . If d y d x = 1 2 d d x sin - 1 f x and y 3 = π 6 , then y - 3 is equal to:

Step-by-step solution

Let t a n - 1 x = θ ⇒ x = t a n θ   ⇒ s i n θ = x 1 + x 2 So, y = x 1 + x 2 + 1 1 + x 2 2 - 1 ⇒ y = x + 1 2 1 + x 2 - 1 ⇒ y = 2 x 1 + x 2 = f x Now, d y d x = 1 2 1 1 - f 2 × f ' x = 1 2 1 1 - 4 x 2 1 + x 2 2 f x = 1 + x 2 2 x 2 - 1 f ' x = 1 + x 2 2 x 2 - 1 × 2 1 + x 2 - 2 x 2 1 + x 2 2 ⇒ d y d x = 1 - x 2 x 2 - 1 1 + x 2 ⇒ d y = 1 - x 2 x 2 - 1 1 + x 2 d x Integrating both sides with respect to x ∫ d y = ∫ 1 - x 2 x 2 - 1 1 + x 2 d x &