Question

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

Answer 1

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

Take $x=\tan \theta$

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

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

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

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

$\Rightarrow \pi -2\theta +\frac{1}{2}\times 2\theta =\frac{2\pi }{3}$

$\Rightarrow \pi -2\theta +\theta =\frac{2\pi }{3}$

$\Rightarrow -\theta =\frac{2\pi }{3}-\pi$

$\Rightarrow -\theta =\frac{2\pi -3\pi }{3}$

$\Rightarrow -\theta =\frac{-\pi }{3}$

$\therefore \theta =\frac{\pi }{3}$

We have $x=\tan \theta =\tan \frac{\pi }{3}=\sqrt{3}$

$\therefore x=\sqrt{3}$