Prove the identity: $\dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} = {\tan^4 \alpha}$
$\begin{aligned}
LHS & = \dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} \\\\
& = \dfrac{3 - 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha}{3 + 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha} \\\\
& = \dfrac{4 - 4 \cos 2 \alpha - 2 \sin^2 2 \alpha}{4 + 4 \cos 2 \alpha - 2 \sin^2 2 \alpha} \\\\
& = \dfrac{2 \left(1 - \cos 2 \alpha\right) - \sin^2 2 \alpha}{2 \left(1 + \cos 2 \alpha \right) - \sin^2 2 \alpha} \\\\
& = \dfrac{2 \times 2 \sin^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2}{2 \times 2 \cos^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2} \\\\
& = \dfrac{4 \sin^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha}{4 \cos^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha} \\\\
& = \dfrac{\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha \cos^2 \alpha} \\\\
& = \dfrac{\sin^2 \alpha \left(1 - \cos^2 \alpha\right)}{\cos^2 \alpha \left(1 - \sin^2 \alpha\right)} \\\\
& = \dfrac{\sin^2 \alpha \times \sin^2 \alpha}{\cos^2 \alpha \times \cos^2 \alpha} \\\\
& = \dfrac{\sin^4 \alpha}{\cos^4 \alpha} \\\\
& = \tan^4 \alpha = RHS
\end{aligned}$
Hence proved.