Prove the identity: $\cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha = 8 \cot 8 \alpha$
$\begin{aligned}
LHS & = \cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha \\\\
& = \dfrac{\cos \alpha}{\sin \alpha} - \dfrac{\sin \alpha}{\cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{\cos 2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{2 \cos 2 \alpha}{2 \sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{2 \cos 2 \alpha}{\sin 2 \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{2 \left(\cos^2 2 \alpha - \sin^2 2 \alpha\right)}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{2 \cos 4 \alpha}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{4 \cos 4 \alpha}{2 \sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{4 \cos 4 \alpha}{\sin 4 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\
& = \dfrac{4 \left(\cos^2 4 \alpha - \sin^2 4 \alpha\right)}{\sin 4 \alpha \cos 4 \alpha} \\\\
& = \dfrac{4 \cos 8 \alpha}{\sin 4 \alpha \cos 4 \alpha} \\\\
& = \dfrac{8 \cos 8 \alpha}{2 \sin 4 \alpha \cos 4 \alpha} \\\\
& = \dfrac{8 \cos 8 \alpha}{\sin 8 \alpha} \\\\
& = 8 \cot 8 \alpha = RHS
\end{aligned}$
Hence proved.