# If tan x + tan

Question:

If $\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{2 \pi}{3}\right)=3$, then prove that $\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}=1$.

Solution:

Given:

$\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(\mathrm{x}+\frac{2 \pi}{3}\right)=3$

$\Rightarrow \tan x+\frac{\tan x+\tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}}+\frac{\tan x+\tan \frac{2 \pi}{3}}{1-\tan x \tan \frac{2 \pi}{3}}=3$

$\Rightarrow \tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3} \tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3} \tan x}=3 \quad\left[\tan 120^{\circ}=-\sqrt{3}\right]$

$\Rightarrow \frac{\tan x\left(1-3 \tan ^{2} x\right)+\tan x+\sqrt{3}+\sqrt{3} \tan ^{2} x+3 \tan x+\tan x-\sqrt{3}-\sqrt{3} \tan ^{2} x+3 \tan x}{1-3 \tan ^{2} x}=3$

$\Rightarrow \frac{9 \tan x-3 \tan ^{3} x}{1-3 \tan ^{2} x}=3$

$\Rightarrow \frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}=1$

Hence proved.