Differentiate the function $\tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}, \quad -\frac{\pi}{4} < x < \frac{\pi}{4}$ with respect to $x$.

$\frac{1}{2}$

The correct answer is Option (1) → $\frac{1}{2}$ ##

Let $y = \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \tan^{-1} \sqrt{\frac{1 - 1 + 2\sin^2 \frac{x}{2}}{1 + 2\cos^2 \frac{x}{2} - 1}} = \tan^{-1} \left( \tan \frac{x}{2} \right) = \frac{x}{2}$

$\left[ ∵\cos x = 1 - 2\sin^2 \frac{x}{2} = 2\cos^2 \frac{x}{2} - 1 \right]$

On differentiating w.r.t. $x$, we get, $\frac{dy}{dx} = \frac{1}{2}$