Explanation

(d)
Given, $\tan^{- 1}{x + 2\cot^{- 1}{x = \frac{2\pi}{3}}}$
$\therefore\ \ \tan^{- 1}{x + 2\tan^{- 1}{\frac{1}{x} = \frac{2\pi}{3}}}$
$\Rightarrow \ \ \tan^{- 1}{x + \tan^{- 1}{\left( \frac{2\left( \frac{1}{x} \right)}{1 - \left( \frac{1}{x} \right)^{2}} \right) = \frac{2\pi}{3}}}$
$\Rightarrow \ \tan^{- 1}{x + \tan^{- 1}\left( \frac{2x}{x^{2} - 1} \right)} = \frac{2\pi}{3}$
$\Rightarrow \ \ \tan^{- 1}\left( \frac{x + \frac{2x}{x^{2} - 1}}{1 - \frac{2x^{2}}{x^{2} - 1}} \right) = \frac{2\pi}{3}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \frac{x(x^{2} + 1)}{- 1(x^{2} + 1)} = - \sqrt{3}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \sqrt{3}$