# Trigonometric Identities

• Oct 6th 2009, 07:35 AM
BeckyDWood
Trigonometric Identities
Prove the identity tan 2x - tan x = tan x sec 2x

Thank you for your time and help
• Oct 6th 2009, 10:09 AM
Amer
Quote:

Originally Posted by BeckyDWood
Prove the identity tan 2x - tan x = tan x sec 2x

Thank you for your time and help

$\displaystyle \tan 2x -\tan x = \tan x \sec 2x$

take the left side

$\displaystyle \tan 2x - \tan x = \frac{\sin 2x }{\cos 2x} - \frac{\sin x }{\cos x}$

$\displaystyle \tan 2x - \tan x = \frac{\sin 2x \cos x - \sin x \cos 2x }{\cos 2x \cos x}$

$\displaystyle \tan 2x - \tan x = \frac{2 \sin x \cos ^2 x - \sin x \cos 2x }{\cos x \cos 2x }$

$\displaystyle \tan 2x - \tan x = \frac{\sin x( 2\cos ^2 x - \cos 2x }{\cos 2x \cos x}$

$\displaystyle \tan 2x - \tan x = \frac{\sin x }{\cos 2x \cos x } =\tan x \sec 2x$
• Oct 6th 2009, 11:02 AM
Krizalid
Quote:

Originally Posted by Amer
$\displaystyle \tan 2x - \tan x = \frac{\sin 2x \cos x - \sin x \cos 2x }{\cos 2x \cos x}$

note that $\displaystyle \sin 2x \cos x - \sin x \cos 2x=\sin(2x-x)=\sin x.$