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