2016/020)Solve $\cos^{-1}\frac{x^2-1}{x^2+1} +\frac{1}{2} \tan^{-1}\frac{-2x}{1-x^2}=\frac{2\pi}{3}$

we have $\tan (2y) = \frac{2\tan y }{1-\tan^2 y}$
hence   $\tan (- 2y) = \frac{-2\tan y }{1-\tan^2 y}$
hence if $x = \tan y$
 $-2y = \tan ^-1 \frac{-2x}{1-x^2}$
or  $\tan ^{-1} \frac{-2x}{1-x^2}= -2 \tan ^{-1} x\cdots(1)$
further $\cos(2y) = \sin ^2 y - \cos^2 y = \frac{\sin ^2 y - \cos^2 y}{\sin ^2 y + \cos^2 y}$
or $\cos(2y) = \frac{\tan  ^2 y - 1}{\tan  ^2 y + 1}$
or $\cos(\pi - 2y) = \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
or $\pi - 2y = \cos ^{-1} \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
putting $\tan y = x$ we get  $\pi - 2 \tan ^{-1} x = \cos ^{-1} \frac{1 - x^2}{1 + x^2}\cdots(2)$
putting values from (1) and (2) in given equation we get
$\pi - 2\tan^{-1} x - (1/2)*2\tan^{-1} x = \frac{2\pi}{3}$
or  $3\tan^{-1}(x) = \frac{\pi}{3}$
or  $\tan^{-1} (x) = \frac{\pi}{9}$
Thus $x = \tan\frac{\pi}{9}$