cot x + cot
Question:

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

Solution:

$\frac{\pi}{3}=60^{\circ}, \frac{2 \pi}{3}=120^{\circ}$

$\mathrm{LHS}=\cot x+\cot \left(60^{\circ}+x\right)+\cot \left(120^{\circ}+x\right)$

$=\cot x+\cot \left(60^{\circ}+x\right)-\cot \left[180^{\circ}-\left(120^{\circ}+x\right)\right]$

$\left(\because-\cot \theta=\cot \left(180^{\circ}-\theta\right)\right)$

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

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

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

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

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

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

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

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

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

$=3 \cot 3 x$

= RHS

Hence proved.