Trigonometry challenging question

$\begin{aligned} X & = \tan x + 2 \tan 2x + 4 \tan 4x + 8 \cot 8x \\ & = \tan x + 2 \tan 2x + 4 \tan 4x + \frac {4 (1-\tan^2 4x)}{\tan 4x} \\ & = \tan x + 2 \tan 2x + \frac 4{\tan 4x} \\ & = \tan x + 2 \tan 2x + \frac {2(1-\tan^2 2x)}{\tan 2x} \\ & = \tan x + \frac 2{\tan 2x} \\ & = \tan x + \frac {1-\tan^2 x}{\tan x} \\ & = \frac 1{\tan x} = \boxed{\cot x} \end{aligned}$

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Generalization: For positive integer $n$ :

$\tan x + 2 \tan 2x + 4 \tan 4x + \cdots + 2^{n-1} \tan 2^{n-1} x + 2^n \cot 2^n x = \cot x$

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