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

We use half-angle identities to express everything in terms of $\frac{x}{2}$:

$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$

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

This follows because $\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 = \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} + \sin^2\frac{x}{2} = 1 + \sin x$

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

$= \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}$

Dividing numerator and denominator by $\cos\frac{x}{2}$:

$= \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}}$

Using the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ with $A = \frac{\pi}{4}$ and $B = \frac{x}{2}$, and noting $\tan\frac{\pi}{4} = 1$:

$\frac{1 - \tan\frac{x}{2}}{1 + 1\cdot\tan\frac{x}{2}} = \tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right)$

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

Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we get $0 < \frac{\pi}{4} - \frac{x}{2} < \frac{\pi}{2}$, which lies within the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of $\tan^{-1}$, so $\tan^{-1}$ and $\tan$ cancel directly.

$= \dfrac{\pi}{4} - \dfrac{x}{2}$