Explanation

(d)
Given, $\tan\left( \frac{x}{2} \right) = cosec\ x - \sin x$
$= \frac{1 + \tan^{2}\frac{x}{2}}{2\tan\frac{x}{2}} - \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}$
$= \frac{\left( 1 + \tan^{2}\frac{x}{2} \right) - 4\tan^{2}\frac{x}{2}}{2\tan\frac{x}{2}\left( 1 + \tan^{2}\frac{x}{2} \right)}$
$\Rightarrow \ 2\tan^{2}\left( \frac{x}{2} \right)\left( 1 + \tan^{2}\frac{x}{2} \right) = \left( 1 - \tan^{2}\frac{x}{2} \right)^{2}$
$\Rightarrow \ \tan^{4}{\frac{x}{2} + 4\tan^{2}{\frac{x}{2} - 1 = 0}}$
$\Rightarrow \tan^{2}{\frac{x}{2} = \frac{- 4 \pm \sqrt{16 + 4}}{2 \times 1}} = - 2 \pm \sqrt{5}$
$\Rightarrow \tan^{2}{\frac{x}{2} = - 2 + \sqrt{5}}$
$\left( \because\tan^{2}{\frac{x}{2} \neq - 2 - \sqrt{5}} \right)$