\(\begin{array}{l}
\quad \dfrac{1}{1+\cos x} - \dfrac{1}{\sin^2x}\\
= \dfrac{\sin^2x - (1 + \cos x)}{(1+\cos x)\sin^2x}\\
= \dfrac{-(1-\sin^2x) - \cos x}{(1+\cos x)\sin^2x}\\
= \dfrac{-\cos^2x - \cos x}{(1+\cos x)\sin^2x}\\
= \dfrac{-\cos x(1 + \cos x)}{(1+\cos x)\sin^2x}\\
= \dfrac{-\cos x}{\sin^2x}\\
= \dfrac{-\dfrac{\cos x}{\sin x}}{\sin x}\\
= \dfrac{-\cot x}{\sin x}
\end{array}\)
