Question

Rewrite 2tan3x in terms of tanx

Answer 1

$2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}$

Use the same identity to expand $\tan2x$.

$\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}$

$\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}$
$2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}$
$2\tan3x=\dfrac{2\tan x\left(2+1-\tan^2x\right)}{1-\tan^2x-2\tan^2x}$
$2\tan3x=\dfrac{2\tan x(3-\tan^2x)}{1-3\tan^2x}$