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:

Let $$f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^2-1, $$ $\displaystyle |x|>1$. If $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2} \frac{\mathrm{~d}}{\mathrm{~d} x}\left(\sin ^{-1}(f(x))\right)$ and $\displaystyle y(\sqrt{3})=\frac{\pi}{6}$, then $\displaystyle y(-\sqrt{3})$ is equal to :