Trigonometric Identities

**BeckyDWood** · October 6th 2009, 07:35 AM

Prove the identity tan 2x - tan x = tan x sec 2x

Thank you for your time and help

**Amer** · October 6th 2009, 10:09 AM

Originally Posted by BeckyDWood

Prove the identity tan 2x - tan x = tan x sec 2x

Thank you for your time and help

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

take the left side

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

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

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

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

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

**Krizalid** · October 6th 2009, 11:02 AM

Originally Posted by Amer

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

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

Thread: Trigonometric Identities

LinkBack

Thread Tools

Display

Trigonometric Identities

Similar Math Help Forum Discussions

Trigonometric Identities

Trigonometric identities

trigonometric identities

trigonometric identities

trigonometric identities

Search Tags