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

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

The correct answer is Option (3) → $\frac{\pi}{4} + \frac{x}{2}$ ##

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

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

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

$= \tan^{-1} \left[ \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right] \,\,\, \left[ \text{Dividing Nr. \& Dr. by } \cos \frac{x}{2} \right]$

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

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

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