Simplify
\[\tan x + 2 \tan 2x + 4 \tan 4x + 8 \cot 8x.\]The answer will be a trigonometric function of some simple function of $x,$ like "$\cos 2x$" or "$\sin (x^3)$".
So we want to simplify $\tan x + 2 \tan 2x + 4 \tan 4x + 8 \cot 8x$. And we should start by changing the $\cot$ to a $\tan$ since the other terms are all $\tan$ terms. Right. So we want to simplify $\tan x + 2 \tan 2x + 4 \tan 4x + 8 \frac{1}{\tan 8x}$. And that means the expression we want to simplify is $\tan x + 2 \tan 2x + 4 \tan 4x + 8 \frac{\cos 8x}{\sin 8x}$. So we want to simplify $\tan x + 2 \tan 2x + 4 \tan 4x + 8 \frac{1}{\tan 8x} - \cot x + \cot x$. Now we can use the identity $\tan a - \cot a = -2 \cot 2a$ to simplify. So the expression simplifies to $-2 \cot 2x + 2 \tan 2x + 4 \tan 4x + 8 \cot 8x + \cot x$. Continuing to use the identity $\tan a - \cot a = -2 \cot 2a$, we can simplify further to $-4 \cot 4x + 4 \tan 4x + 8 \cot 8x + \cot x$. And now we can simplify to just $\cot x$. Right, so $\tan x + 2 \tan 2x + 4 \tan 4x + 8 \cot 8x = \cot x$.

