Explanation

(b)
Given, $\tan^{- 1}{\frac{1 - x}{1 + x} = \frac{1}{2}\tan^{- 1}x}$
Let $x = \tan\theta$
$\therefore\ \tan^{- 1}\left( \frac{1 - \tan\theta}{1 + \tan\theta} \right) = \frac{1}{2}\tan^{- 1}{(\tan\theta)}$
$\Rightarrow \tan^{- 1}\left\{ \tan\left( \frac{\pi}{4} - \theta \right) \right\} = \frac{1}{2}\tan^{- 1}{(\tan\theta)}$
$\Rightarrow \ \ \ \frac{\pi}{4} - \theta = \frac{\theta}{2}\ \Rightarrow \ \theta = \frac{\pi}{6}$
$\therefore\ \ \ x = \tan{\frac{\pi}{6} = \frac{1}{\sqrt{3}}}$