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