# Thread: Identity problem: dont understand Tan3x

1. ## Identity problem: dont understand Tan3x

Prove the identity:
23.) (tan3x - tanx)/(1 + tan3xtanx) = 2tanx/(1-tan^2 x) <--- tan squared

elp!!!

Prove the identity:
23.) (tan3x - tanx)/(1 + tan3xtanx) = 2tanx/(1-tan^2 x) <--- tan squared

elp!!!
$\displaystyle \frac {\tan3x - tanx}{1+\tan3x tanx}$

Use the trig identity:
$\displaystyle tan(A-B) = \frac {tanA - tanB}{1+ tanA tanB}$

$\displaystyle = \tan (3x-x)$

$\displaystyle = \tan 2x$

Use the trig identity:
$\displaystyle tan2A= \frac {2tanA}{1 - \tan^2A}$

$\displaystyle = \frac{2 tanx}{1- \tan^2x}$

We also can do it "the long way" . . .

We're expected to know that: . $\displaystyle \tan2x \;=\;\frac{2\tan x}{1 - \tan^2x}\qquad\tan3x \:=\:\frac{\tan2x + \tan x}{1 - \tan2x\tan x}$

Prove the identity: $\displaystyle \frac{\tan3x - \tan x}{1 + \tan3x\tan x} \:= \:\frac{2\tan x}{1 -\tan^2\!x}$

The left side is: .$\displaystyle \displaystyle{\frac{\left(\frac{\tan 2x + \tan x}{1 - \tan2x\tan x}\right) - \tan x}{1 + \left(\frac{\tan 2x + \tan x}{1 - \tan2x\tan x}\right)\tan x}}$

Multiply top and bottom by $\displaystyle 1 - \tan2x\tan x:$

. . $\displaystyle \frac{\tan 2x + \tan x - \tan x(1 - \tan 2x\tan x)}{1 - \tan2x\tan x + (\tan2x + \tan x)\tan x}$

We have: .$\displaystyle \frac{\tan2x + \tan x - \tan x + \tan2x\tan^2\!x}{1 - \tan2x\tan x + \tan2x\tan x + \tan^2x}$ $\displaystyle = \;\frac{\tan2x + \tan2x\tan^2\!x}{1 + \tan^2x}$

Factor: . $\displaystyle \frac{\tan2x(1 + \tan^2\!x)}{1 + \tan^2\!x} \;=\;\tan2x \;= \;\frac{2\tan x}{1 - \tan^2\!x}$