Proving the identity $\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$ by considering a more general form

Defining $t:=\tan y$, $$2\cot 2y+\tan y-\cot y=\frac{1-t^2}{t}+t-\frac{1}{t}=0.$$This will work for any $y$, so I guess I've answered your questions in reverse order (but to both I say yes).

Hint:

$$\cot y-\tan y=\dfrac{\cos^2y-\sin^2y}{\cos y\sin y}=\dfrac{\cos2y}{\dfrac{\sin2y}2}=?$$