# How does one verify (secx-cosx)^2=tan^2x-sin^2x?

## ${\left(\sec x - \cos x\right)}^{2} = {\tan}^{2} x - {\sin}^{2} x$

Apr 23, 2018

See below.

#### Explanation:

We'll leave the right side alone and attempt to get the left side to match it.

Expand the left side:

$\left(\sec x - \cos x\right) \left(\sec x - \cos x\right) = {\tan}^{2} x - {\sin}^{2} x$

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

$2 \sec x \cos x = \frac{2}{\cancel{\cos}} x \cancel{\cos} x = 2 ,$ so we get

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

Rewrite, splitting up the $- 2$ into $- 1 - 1$:

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

Recall the following identities:

${\sec}^{2} x - 1 = {\tan}^{2} x$
${\sin}^{2} x + {\cos}^{2} x = 1 \to {\cos}^{2} x - 1 = - {\sin}^{2} x$

So, we get

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