d d x tan − 1 ⁡ cos ⁡ x 1 + sin ⁡ x =

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

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

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