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