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?
Prove $\cot(x) +\cot(\frac{\pi}{3}+x) + \cot(\frac{2\pi}{3}+x) = \frac{3-9\tan^2x}{3\tan x-\tan^3x}$
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probably you made mistake somewhere – S L May 02 '14 at 09:33
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You can tell your simplification of the LHS must be incorrect by plugging in $x=0$: $\tan0=0$, so the expression you got reduces to $1$, but the expression you're aiming at is infinite at $x=0$. – Barry Cipra May 02 '14 at 19:00
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Yeah there's a mistake in my simplification. :) – Harshal Gajjar May 03 '14 at 13:24
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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 $(1)$)
lab bhattacharjee
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See also : http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series and http://math.stackexchange.com/questions/218766/prove-the-trigonometric-identity-35 – lab bhattacharjee May 02 '14 at 18:55