2016/067) Solve $\cos^{ -1}\frac{x^2-1}{x^2+1} + \frac{1}{2}\ tan^{-1}\frac{-2x}{1-x^2}=\frac{2\pi}{3}$

Let $x=\tan \,t$
we know $\cos^{-1}-x = \pi - \cos^{-1}x$
hence $\cos^{-1}\frac{x^2 - 1}{x^2 + 1} = \pi - \cos^{-1}\frac{1 - x^2}{x^2 + 1}$
$= \pi - \cos^{-1}\frac{1 - \tan 2 t}{1 + \tan  ^2 t}= \pi - 2t$

Similarly, $\tan^{-1}\frac{-2x}{1 - x^2} = -\tan^{-1}\frac{2x}{1 - x^2} = - 2t$

The given equation reduces to:
$\pi - 2t - \frac{1}{2}2t= \frac{2\pi}{3}$
$3tan^-1(x) = \frac{\pi}{3}$
or $\tan^{-1}x = \frac{\pi}{9}$
Thus $x = \tan\frac{\pi}{9}$