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