Solve the equation $1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$ How to solve this? Any advice?
$$1-\tan x + \tan^2 x - \tan^3 x + ... = \frac{\tan 2x}{1+\tan2x}$$
Next step I do this
$\sum\limits_{n=0}^\mathbb{\infty}(-1)^n  \tan^nx = \frac{\tan 2x}{1+\tan2x} $
But I don't know next step. I am culeless, thanks for any advice.

Notice, for the sum of infinite series on LHS,  $|\tan x|<1$
$$1-\tan x+\tan^2 x-\tan^3 x+\ldots =\frac{\tan 2x}{1+\tan 2x}$$
$$\frac{1}{1-(-\tan x)}=\frac{\frac{2\tan x}{1-\tan^ 2x}}{1+\frac{2\tan x}{1-\tan^ 2x}}$$
$$\frac{1}{1+\tan x}=\frac{2\tan x}{1-\tan^2 x+2\tan x}$$
$$1-\tan^2 x+2\tan x=2\tan x+2\tan^2 x$$
$$3\tan^2 x=1$$$$\tan^2 x=\frac{1}{3}$$
I hope you can solve further
