Express $\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right)$, $-\frac{3\pi}{2} < x < \frac{\pi}{2}$ in the simplest form.

$\frac{\pi}{4} + \frac{x}{2}$

The correct answer is Option (2) → $\frac{\pi}{4} + \frac{x}{2}$ ##

We write

$\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) = \tan^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2\sin \frac{x}{2} \cos \frac{x}{2}} \right]$

$= \tan^{-1} \left[ \frac{\left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)}{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2} \right]$

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

$= \tan^{-1} \left[ \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right] = \frac{\pi}{4} + \frac{x}{2}$