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

$\Rightarrow x=\mathrm{t}\mathrm{a}\mathrm{n}\theta \hspace{0.17em}\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}\theta =\frac{x}{\sqrt{1+x^2}}$

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

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

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

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

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

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

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

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

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

Integrating both sides with respect to $x$

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

$\Rightarrow y=-{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}x+c$

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

$y=-{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}x+\frac{\pi }{2}={\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}x$

Now, $y\left(-\sqrt{3}\right)={\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}x\left(-\sqrt{3}\right)=\frac{5\pi }{6}$