If $x>1and{\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=\frac{2\pi }{3}$, then $x=$

${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)=\pi -{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\pi -2{\tan }^{-1}x(asx>1)$

${\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=-{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\pi -2{\tan }^{-1}x(asx>1)$

$\therefore$Given equation is $2\pi -4{\tan }^{-1}x=\frac{2\pi }{3}$

$\therefore {\tan }^{-1}x=\frac{\pi }{3}$

$\therefore x=\sqrt{3}$