Explanation

(a)
Given, $\tan^{- 1}{\left( \frac{1}{\sqrt{\cos\alpha}} \right) - \tan^{- 1}{\left( \sqrt{\cos\alpha} \right) = x}}$
$\Rightarrow \ \ \tan^{- 1}{\left( \frac{\frac{1}{\sqrt{\cos\alpha}} - \sqrt{\cos\alpha}}{1 + \frac{1}{\sqrt{\cos\alpha}}.\sqrt{\cos\alpha}} \right) = x}$
$\Rightarrow \ \ \ \frac{1 - \cos\alpha}{2\sqrt{\cos\alpha}} = \tan x$
$\Rightarrow \ \ \frac{2\sqrt{\cos\alpha}}{1 - \cos\alpha} = \cot x$
$\Rightarrow \ \ cosec\ x = \frac{1 + \cos\alpha}{1 - \cos\alpha}$
$\Rightarrow \ \sin{x = \frac{1 - \cos\alpha}{1 + \cos\alpha}}$
$\Rightarrow \ \sin{x = \frac{2\sin^{2}\left( \frac{\alpha}{2} \right)}{2\cos^{2}\left( \frac{\alpha}{2} \right)}} = \tan^{2}\left( \frac{\alpha}{2} \right)$