A problem by Clent Pacilan

$\begin{aligned} \dfrac{\sec x \sin^2 x}{1 + \sec x} & = \dfrac{\frac 1{\cos x} (1 - \cos^2 x)}{1 + \frac 1{\cos x}} ~~~~~~ [1-\cos^2 x = \sin^2 x] \\ \large & = \dfrac{\frac{1 - \cos^2 x}{\cos x}}{\frac{1+ \cos x}{\cos x}} \\ & = \dfrac{(1 - \cos^2x)(\cos x)}{(1+\cos x)(\cos x)} \\ & = \dfrac{1 - \cos^2 x}{1+\cos x} \\ & = \dfrac{(1+\cos x)(1-\cos x)}{(1+\cos x)} \\ & = \boxed{1 - \cos x} \end{aligned}$