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