The simplified form of $\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right), -\frac{\pi}{2}<x<\frac{\pi}{2}$ is

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

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

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

Multiply numerator and denominator inside $\tan^{-1}$ by $(1 - \sin x)$:

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

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

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

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

This is equal to $\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)$ using the identity:

$\frac{1 - \sin x}{\cos x} = \cot\left(\frac{x}{2}\right)$

Now, use the identity: $\cot\left(\frac{x}{2}\right) = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$

So, $\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right)$

Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, the angle $\frac{\pi}{4} - \frac{x}{2}$ lies in principal domain

Therefore, final answer is:

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