Prove that

Submitted by alpha on Thu, 07/02/2009 - 17:34.

if $tanx+tan[{x+\frac{\pi}{3}}]+tan[{x+\frac{2\pi}{3}}]=3$
Prove that $\frac{3tanx-tan^3x }{1-3tan^2x }=1$

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

$\tan x=t$

$t+\frac{t+\sqrt3}{1-\sqrt3t}+\frac{t-\sqrt3}{1+\sqrt3t}=\frac{9t-3t^3}{1-3t^2}=3$