$\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{\pi}{3}-x\right)=3 \cot 3 x$
$\frac{\pi}{3}=60^{\circ}$
$\mathrm{LHS}=\cot x+\cot \left(60^{\circ}+x\right)-\cot \left(60^{\circ}-x\right)$
$=\frac{1}{\tan x}+\frac{1}{\tan \left(60^{\circ}+x\right)}-\frac{1}{\tan \left(60^{\circ}-x\right)}$
$=\frac{1}{\tan x}+\frac{1-\sqrt{3} \tan x}{\sqrt{3}+\tan x}-\frac{1+\sqrt{3} \tan x}{\sqrt{3}-\tan x}$
$\left[\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right.$ and $\left.\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]$
$=\frac{1}{\tan x}-\frac{8 \tan x}{3-\tan ^{2} x}$
$=\frac{3-\tan ^{2} x-8 \tan ^{2} x}{(\tan x)\left(3-\tan ^{2} x\right)}$
$=\frac{3-9 \tan ^{2} x}{3 \tan x-\tan ^{3} x}$
$=3\left(\frac{1-3 \tan ^{2} x}{3 \tan x-\tan ^{3} x}\right)$
$=3 \times \frac{1}{\tan 3 x} \quad\left(\because \tan 3 \theta=\frac{1-3 \tan ^{2} \theta}{3 \tan \theta-\tan ^{3} \theta}\right)$
$=3 \cot 3 x$
= RHS
Hence proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.