Question

Solve the following equation
${\cos }^{-1}\frac{x^2-1}{x^2+1}+{\tan }^{-1}\frac{2x}{x^2-1}=\frac{2\pi }{3}$.

Answer 1

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

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

$[\because {\cos }^{-1}(-x)=\pi -{\cos }^{-1}x{\tan }^{-1}(-x)=-{\tan }^{-1}x]$

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

$\Rightarrow 4{\tan }^{-1}x=\pi -\frac{2\pi }{3}$

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

$\Rightarrow {\tan }^{-1}x=\frac{\pi }{12}\Rightarrow x=\tan \frac{\pi }{12}\Rightarrow x=\tan \left(\frac{\pi }{12}\right)$