Prove the identity: $\dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right)$
$\begin{aligned}
\cot \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{\tan \left(\dfrac{\pi}{4} + \alpha\right)} \\\\
& = \dfrac{1 - \tan \left(\dfrac{\pi}{4}\right) \tan \alpha}{\tan \dfrac{\pi}{4} + \tan \alpha} \\\\
& = \dfrac{1 - \tan \alpha}{1 + \tan \alpha} \\\\
& = \dfrac{1 - \dfrac{\sin \alpha}{\cos \alpha}}{1 + \dfrac{\sin \alpha}{\cos \alpha}} \\\\
& = \dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}
\end{aligned}$
$\begin{aligned}
\therefore \; RHS & = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right) \\\\
& = \left(\dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2 \\\\
& = \dfrac{\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha}{\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha} \\\\
& = \dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = LHS
\end{aligned}$
Hence proved.