Trigonometric Identities

Quote:

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

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

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

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

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

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