# How do you prove sec x - cos x = sin x tan x?

Nov 13, 2015

See explanation...

#### Explanation:

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

So:

$\sec \left(x\right) - \cos \left(x\right)$

$= \frac{1}{\cos \left(x\right)} - \cos \left(x\right)$

$= \frac{1}{\cos \left(x\right)} - {\cos}^{2} \frac{x}{\cos} \left(x\right)$

$= \frac{1 - {\cos}^{2} \left(x\right)}{\cos} \left(x\right)$

$= \frac{{\sin}^{2} \left(x\right)}{\cos} \left(x\right)$

$= \sin \left(x\right) \sin \frac{x}{\cos} \left(x\right)$

$= \sin \left(x\right) \tan \left(x\right)$