# tan x

Question:

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

Solution:

$\frac{\pi}{3}=60^{\circ}$

LHS $=\tan x+\tan \left(60^{\circ}+x\right)-\tan \left(60^{\circ}-x\right)$

$=\tan x+\left(\frac{\tan 60^{\circ}+\tan x}{1-\tan 60^{\circ} \tan x}\right)-\left(\frac{\tan 60^{\circ}-\tan x}{1+\tan 60^{\circ} \tan x}\right)$

$\left[\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right.$ and $\left.\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]$

$=\tan x+\frac{\sqrt{3}+\tan x}{1-\sqrt{3} \tan x}-\frac{\sqrt{3}-\tan x}{1+\sqrt{3} \tan x}$

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

$=\tan x+\frac{8 \tan x}{1-3 \tan ^{2} x}$

$=\frac{\tan x-3 \tan ^{3} x+8 \tan x}{1-3 \tan ^{2} x}$

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

$=3\left(\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\right) \quad\left(\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$

$=3 \tan 3 x$

$=\mathrm{RHS}$

Hence proved.