Question

Prove That: tan x + tan(pi/3+x) + tan(2pi/3+x) = 3 tan3x

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$\displaystyle \: tan \: (A + B) = \frac{tan \: A + tan \: B}{1 - tan \: A \: tan B }$

2.

$\displaystyle \: tan \: \frac{2\pi}{3} = tan(\pi - \frac{\pi}{3} ) = - tan \: \frac{\pi}{3} = - \sqrt{3}$

CALCULATION

$\displaystyle \: tan \: x + tan(x + \: \frac{\pi}{3}) + tan(x + \: \frac{2\pi}{3})$

$= \displaystyle \: \: tanx \: + \frac{tan \: x + tan \: \frac{\pi}{3} }{1 - tan \: x \: tan \frac{\pi}{3} } + \frac{tan \: x + tan \: \frac{2\pi}{3} }{1 - tan \: x \: tan \frac{2\pi}{3} }$

$= \displaystyle \: \: tanx \: + \frac{tan \: x + \sqrt{3} }{1 - \sqrt{3} tan \: x } + \frac{tan \: x - \sqrt{3} }{1 + \sqrt{3} tan \: x \:} \: \: (by \: \: formula \: 2)$

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

$= \: \frac{tanx - 3 {tan}^{3} x + tanx + \sqrt{3 } {tan}^{2} x + \sqrt{3} + 3tanx - \sqrt{3 } {tan}^{2} x - \sqrt{3} +tanx+ 3tanx }{1 - 3 {tan}^{2}x }$

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

$= 3 \: tan \: 3x$