Question

Solve:
${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\tan }^{-1}\frac{2x}{x^2-1}=\frac{2\pi }{3}$

Answer 1

${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)=\frac{2\pi }{3}$

$\Rightarrow \pi -{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)-{\tan }^{-1}\frac{2x}{1-x^2}=\frac{2\pi }{3}$

$\pi -2{\tan }^{-1}x-2{\tan }^{-1}x=\frac{2\pi }{3}$

$4{\tan }^{-1}x=\frac{\pi }{3}$

${\tan }^{-1}x=\frac{\pi }{12}$

$\Rightarrow x=\tan \frac{\pi }{12}$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Hence, the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}.$