Let f(x) = ( ( ^ - 1 x) + ( ^ - 1 x)) ^ 2 - 1 , |x| > 1 . If...

$⇒ x = tan\theta ⇒ sin\theta = \dfrac{x}{\sqrt{1 + {x}^{2}}}$

So, $y = {\left(\dfrac{x}{\sqrt{1 + {x}^{2}}} + \dfrac{1}{\sqrt{1 + {x}^{2}}}\right)}^{2} - 1$

$⇒ y = \dfrac{{\left(x + 1\right)}^{2}}{1 + {x}^{2}} - 1$

$⇒ y = \dfrac{2x}{1 + {x}^{2}} = f\left(x\right)$

Now, $\dfrac{dy}{dx} = \dfrac{1}{2}\dfrac{1}{\sqrt{1 - {f}^{2}}} \times {f}^{'}\left(x\right)$

$= \dfrac{1}{2}\dfrac{1}{\sqrt{1 - \dfrac{4{x}^{2}}{{\left(1 + {x}^{2}\right)}^{2}}}}f\left(x\right)$

$= \dfrac{\left(1 + {x}^{2}\right)}{2\left({x}^{2} - 1\right)}{f}^{'}\left(x\right)$

$= \dfrac{1 + {x}^{2}}{2\left({x}^{2} - 1\right)} \times 2\dfrac{1 + {x}^{2} - 2{x}^{2}}{{\left(1 + {x}^{2}\right)}^{2}}$

$⇒ \dfrac{dy}{dx} = \dfrac{1 - {x}^{2}}{\left({x}^{2} - 1\right)\left(1 + {x}^{2}\right)}$

$⇒ dy = \dfrac{1 - {x}^{2}}{\left({x}^{2} - 1\right)\left(1 + {x}^{2}\right)}dx$

Integrating both sides with respect to $x$

$∫ dy = ∫ \dfrac{1 - {x}^{2}}{\left({x}^{2} - 1\right)\left(1 + {x}^{2}\right)}dx$

$⇒ y = - \tan^{- 1}x + c$

Given, $y\left(\sqrt{3}\right) = \dfrac{\pi}{6} ⇒ \dfrac{\pi}{6} = - \dfrac{\pi}{3} + c ⇒ c = \dfrac{\pi}{2}$

$y = - \tan^{- 1}x + \dfrac{\pi}{2} = \cot^{- 1}x$

Now, $y\left(- \sqrt{3}\right) = \cot^{- 1}x\left(- \sqrt{3}\right) = \dfrac{5\pi}{6}$