Explanation

(c)
We have, $\tan^{- 1}{\frac{x - 1}{x + 2} + \tan^{- 1}\ \frac{x + 1}{x + 2} = \frac{\pi}{4}}$
$\Rightarrow \tan^{- 1}\left\lbrack \frac{\frac{x - 1}{x + 2} + \frac{x + 1}{x + 2}}{1 - \left( \frac{x - 1}{x + 2} \right)\left( \frac{x + 1}{x + 2} \right)} \right\rbrack = \frac{\pi}{4}$
$\Rightarrow \left\lbrack \frac{2x(x + 2)}{x^{2} + 4 + 4x - x^{2} + 1} \right\rbrack = \tan\ \frac{\pi}{4}$
$\Rightarrow \frac{2x(x + 2)}{4x + 5} = 1$
$\Rightarrow 2x^{2} + 4x = 4x + 5$
$\Rightarrow x = \pm \sqrt{\frac{5}{2}}$