Prove the identity: $\dfrac{1 - 2 \cos^2 \alpha}{2 \tan \left(\alpha - \dfrac{\pi}{4}\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)} = 1$
$\begin{aligned}
\tan \left(\alpha - \dfrac{\pi}{4}\right) & = \dfrac{\tan \alpha - \tan \left(\dfrac{\pi}{4}\right)}{1 + \tan \alpha \tan \left(\dfrac{\pi}{4}\right)} \\\\
& = \dfrac{\tan \alpha - 1}{1 + \tan \alpha} \\\\
& = \dfrac{\dfrac{\sin \alpha}{\cos \alpha} - 1}{\dfrac{\sin \alpha}{\cos \alpha} + 1} \\\\
& = \dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\sin \left(\dfrac{\pi}{4} + \alpha\right) & = \sin \left(\dfrac{\pi}{4}\right) \cos \alpha + \cos \left(\dfrac{\pi}{4}\right) \sin \alpha \\\\
& = \dfrac{1}{\sqrt{2}} \times \cos \alpha + \dfrac{1}{\sqrt{2}} \times \sin \alpha \\\\
& = \dfrac{1}{\sqrt{2}} \left(\sin \alpha + \cos \alpha\right)
\end{aligned}$
$\begin{aligned}
\therefore \; \sin^2 \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right) ^2 \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{1 - 2 \cos^2 \alpha}{2 \tan \left(\alpha - \dfrac{\pi}{4}\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)} \\\\
& = \dfrac{- \left(2 \cos^2 \alpha - 1\right)}{2 \left(\dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}\right) \times \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right)^2} \\
& \hspace{2cm} \;\;\; \left[\text{by equations (1) and (2)}\right] \\\\
& = \dfrac{- \cos 2 \alpha}{- \left(\cos \alpha - \sin \alpha\right) \left(\sin \alpha + \cos \alpha\right)} \\\\
& = \dfrac{- \cos 2 \alpha}{- \left(\cos^2 \alpha - \sin^2 \alpha\right)} \\\\
& = \dfrac{- \cos 2 \alpha}{- \cos 2 \alpha} \\\\
& = 1 = RHS
\end{aligned}$
Hence proved.