<s>[INST] 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.
$$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$
 [/INST] This follows easily from the identity:
$$\cot x - 2 \cot 2x = \tan x$$
Which can easily be seen by using $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$ as follows:
$$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin ^2 x}{\sin x \cos x} = 2 \cot 2x$$
We now have
$$\cot \alpha - 2 \cot 2\alpha = \tan \alpha$$
$$2\cot 2\alpha - 4 \cot 4\alpha = 2\tan 2\alpha$$
$$4\cot 2\alpha - 8 \cot 4\alpha = 4\tan 4\alpha$$
Adding gives us the result.
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