$\begin{array}{l} Let f (x) = {(\sin (\tan^{- 1} x) + \sin (\cot^{- 1} x))}^{2} - 1, | x | > 1 . If \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\sin^{- 1} (f (x))) and y (\sqrt{3}) = \frac{π}{6} then \\ y (- \sqrt{3}) is equal to: \end{array}$

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$\frac{π}{3}$

$\frac{5 π}{6}$

$- \frac{π}{6}$

$\frac{2 π}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} Let \tan^{- 1} x = θ θ \in (\frac{- π}{2}, \frac{- π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) as | x | > 1 \\ ∴ f (x) = (\sin θ + \cos θ)^{2} - 1 = \sin 2 θ = \frac{2 x}{1 + x^{2}} \\ Now \sin^{- 1} f (x) = \sin^{- 1} \sin 2 θ = \{\begin{cases} - π - 2 θ, θ \in (\frac{- π}{2}, \frac{π}{4}) \\ π - 2 θ, θ \in (\frac{π}{4}, \frac{π}{2}) \end{cases} \\ ∴ \frac{d}{d x} (\sin^{- 1} f (x)) = \frac{- 1}{1 + x^{2}}, | x | > 1 ∴ \frac{d y}{d x} = \frac{- 1}{1 + x^{2}} \\ \Rightarrow y = \{\begin{matrix} C_{1} - \tan^{- 1} x, x > 1 \\ C_{2} - \tan^{- 1} x, x < - 1 \end{matrix} y (\sqrt{3}) = \frac{π}{6} \Rightarrow C_{1} = \frac{π}{2} \end{array}$