Explanation

(b)
$3\sin^{- 1}{\frac{2x}{1 + x^{2}} - 4\cos^{- 1}\ \frac{1 - x^{2}}{1 + x^{2}} + 2\tan^{- 1}\ \frac{2x}{1 - x^{2}} = \frac{\pi}{3}}$
On putting $x = \tan{\theta,}$ we get
$3\sin^{- 1}{\left( \frac{2\tan\theta}{1 + \tan^{2}\theta} \right) - 4\cos^{- 1}\left( \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} \right)} + 2\tan^{- 1}{\left( \frac{2\tan\theta}{1 - \tan^{2}\theta} \right) = \frac{\pi}{3}}$
$\Rightarrow 3\sin^{- 1}{(\sin{2\theta) - 4\cos^{- 1}{\left( \cos{2\theta} \right) + 2\tan^{- 1}{(\tan{2\theta) = \frac{\pi}{3}}}}}}$
$\Rightarrow 3(2\theta) - 4(2\theta) + 2(2\theta) = \frac{\pi}{3}$
$\Rightarrow 6\theta - 8\theta + 4\theta = \frac{\pi}{3}$
$\Rightarrow \theta = \frac{\pi}{6} \Rightarrow \tan^{- 1}{x = \frac{\pi}{6}}$
$\Rightarrow x = \tan{\frac{\pi}{6} \Rightarrow x = \frac{1}{\sqrt{3}}}$