$\begin{align}  & y=\frac{1}{1+\cos x} \\ & y={{(1+\cos x)}^{-1}} \\ & \text{Let }u=1+\cos x \\ & y={{u}^{-1}},\text{  }\frac{dy}{du}=-{{u}^{-2}} \\ & \frac{du}{dx}=-\sin x \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=-{{u}^{-2}}\times (-\sin x) \\ & \frac{dy}{dx}=\frac{\sin x}{{{u}^{2}}}=\frac{\sin x}{{{(1+\cos x)}^{2}}} \\\end{align}$