Showing the identity: $\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$ My knowledge of trigonometry is still insufficient to resolve this problem. Any help would be greatly appreciated.
$$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$

Hoping this is not homework:
$$\cot x - 8\cot 8x = \cot x -8\cot 8x= \cot x-8\frac{\cot^24x -1}{2\cot 4x}$$
$$= \cot x-4\frac{\cot^24x -1}{\cot 4x}= \cot x -4\cot 4x +4\tan 4x$$
$$= \cot x -4 \frac{\cot^22x -1}{2\cot 2x}+4\tan 4x=\cot x -2\cot 2x + 2\tan 2 x + 4 \tan 4x$$
$$=\cot x -2 \frac{\cot^2x -1}{2\cot x}+2\tan 2x + 4\tan 4x = \cot x - \cot x + \tan x + 2\tan 2x + 4\tan 4x $$
$$=\tan x + 2\tan 2x + 4\tan 4x $$
