Prove $\cot(x) +\cot(\frac{\pi}{3}+x) + \cot(\frac{2\pi}{3}+x) = \frac{3-9\tan^2x}{3\tan x-\tan^3x}$

Question

As the title says. I tried simplifying the LHS but got: $$\frac{\tan^2x+7\tan x- \sqrt{3}}{\tan^2x-\sqrt{3}}$$ What should I do next?

Answer 1

From this, $\displaystyle\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$

$\displaystyle\implies\cot3x=\frac{1-3\tan^2x}{3\tan x-\tan^3x}\ \ \ \ (1)$

Multiplying the numerator & the denominator by $\cot^3x,$

$\displaystyle\cot3x=\frac{\cot^3x-3\cot x}{3\cot^2x-1}\ \ \ \ (2)$

If $\displaystyle\cot3x=\cot3A\iff\tan3x=\tan3A\implies3x=n\pi+3A$ where $n$ is any integer

$\displaystyle\implies x=\frac{n\pi}3+A$ where $n\equiv0,1,2\pmod3$

Usng $\displaystyle(2),\frac{\cot^3x-3\cot x}{3\cot^2x-1}=\cot3x=\cot3A$

$\displaystyle\iff\cot^3x-3\cot3A\cot^2x-3\cot x+\cot3A=0$

Using Vieta's formula, $\displaystyle\sum_{n=0}^2\cot\left(\frac{n\pi}3+A\right)=\frac{3\cot3A}1=3\cdot\frac{1-3\tan^2A}{3\tan A-\tan^3A}$ (using $(1)$)