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.
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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
Harish Chandra Rajpoot
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If one leaves the ratio on the RHS untouched one arrives swiftly at $$\tan x\tan2x=1$$ which has a nice geometrical interpretation allowing to solve it explicitely. – Did Nov 05 '15 at 07:17
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You have to put the restriction that the values of $|\tan x | <1$ for this to work. recall the formula for a sum of an infinite geometric series
cgo
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