\(\begin{array}{l}
\quad \dfrac{\sin^2x}{1 + \cot x} + \dfrac{\cos^2x}{1 + \tan x}\\
= \dfrac{\sin^2x}{1 + \dfrac{\cos x}{\sin x}} + \dfrac{\cos^2x}{1 + \dfrac{\sin x}{\cos x}}\\
= \dfrac{\sin^2x}{\dfrac{\sin x + \cos x}{\sin x}} + \dfrac{\cos^2x}{\dfrac{\sin x + \cos x}{\cos x}}\\
= \dfrac{\sin^3x}{\sin x + \cos x } + \dfrac{\cos^3x}{\sin x + \cos x}\\
= \dfrac{\sin^3x+ \cos^3x}{\sin x + \cos x}\\
= \dfrac{(\sin x + \cos x)(\sin^2x - \sin x\cos x + \cos^2x)}{\sin x + \cos x}\\
= \sin^2x + \cos^2x - \sin x\cos x\\
= 1 - \sin x\cos x
\end{array}\)
