Explanation

(b)
Let $y = \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - x}{1 + x}} \right\}}$
Put $x = \cos{\theta\ \Rightarrow \theta = \cos^{- 1}x}$
$\Rightarrow \ y = \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \right\}}$
$\Rightarrow y = \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}} \right\}}$
$= \sin^{2}{\cot^{- 1}\left( \tan\frac{\theta}{2} \right)}$
$\Rightarrow y = \sin^{3}\left( \frac{\pi}{2} - \frac{\theta}{2} \right)$
On differentiating w.r.t. $\theta$, we get
$\frac{dy}{d\theta} = 2\sin\ \left( \frac{\pi}{2} - \frac{\theta}{2} \right)\cos\ \left( \frac{\pi}{2} - \frac{\theta}{2} \right) - \left( - \frac{1}{2} \right)$
$\Rightarrow \frac{dy}{d\theta} = - \frac{\sin{(\pi - \theta)}}{2} = - \frac{\sin\theta}{2} = - \frac{1}{2}\sqrt{1 - x^{2}}$
$\therefore\frac{dy}{dx} = \frac{dy}{d\theta}.\frac{d\theta}{dx} = \frac{- 1}{2}\sqrt{1 - x^{2}}\frac{d}{dx}(\cos^{- 1}{x)}$
$= - \frac{\sqrt{1 - x^{2}}}{2}\left( \frac{- 1}{\sqrt{1 - x^{2}}} \right) = \frac{1}{2}$