Solution of the equation ${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\sin }^{-1}\left(\frac{2x}{x^2+1}\right)+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=\frac{2\pi }{3}$ is / are

$\begin{array}{r}L.H.S={\cos }^{-1}\left(-\frac{1-x^2}{1+x^2}\right)+{\sin }^{-1}\frac{2x}{1+x^2}+{\tan }^{-1}\left(-\frac{2x}{1-x^2}\right)\\ =\pi -{\cos }^{-1}\frac{1-x^2}{1+x^2}+{\sin }^{-1}\frac{2x}{1+x^2}-{\tan }^{-1}\frac{2x}{1-x^2}\end{array}$
Case I : x< -1
$=\pi +2{\tan }^{-1}x+\left(-\pi -2{\tan }^{-1}x\right)+\left(\pi +2{\tan }^{-1}x\right)$
$\begin{array}{l}=\frac{2\pi }{3}\\ \Rightarrow 2{\tan }^{-1}x=\frac{2\pi }{3}-\pi =-\frac{\pi }{3}\Rightarrow x=\tan \left(-\frac{\pi }{6}\right)=-\frac{1}{\sqrt{3}}\\ \text{ Since }-\frac{1}{\sqrt{3}}>-1\text{ so }x=-\frac{1}{\sqrt{3}}\text{ is not admissible. }\end{array}$
Case II :-1 < x <0
$=\pi +2{\tan }^{-1}x+2{\tan }^{-1}x-2{\tan }^{-1}x=\frac{2\pi }{3}$
$\Rightarrow {\tan }^{-1}x=-\frac{\pi }{6}\Rightarrow x=-\tan \frac{\pi }{6}=-\frac{1}{\sqrt{3}}\text{ (Valid) }$
Case III : 0 < x < 1
$=\pi -2{\tan }^{-1}x+2{\tan }^{-1}x-2{\tan }^{-1}x=\frac{2\pi }{3}$
$\Rightarrow {\tan }^{-1}x=\frac{\pi }{6}\Rightarrow x=\frac{1}{\sqrt{3}}\left(\mathrm{valid}\right)$
Case IV : x>1
$\begin{array}{l}\text{ L.H.S. }=\pi -2{\tan }^{-1}x+\pi -2{\tan }^{-1}x+\pi -2{\tan }^{-1}x=\frac{2\pi }{3}\\ \Rightarrow -6{\tan }^{-1}x=\frac{2\pi }{3}-3\pi =-\frac{7\pi }{3}\\ \Rightarrow {\tan }^{-1}x=\frac{7\pi }{18}\Rightarrow x=\tan \left(\frac{\pi }{2}-\frac{\pi }{9}\right)=\cot \frac{\pi }{9}=\cot {20}^{\circ }>1\end{array}$