# How do you verify Sec^4x-tan^4x=sec^2x+tan^2x?

Sep 13, 2016

Prove trig expression

#### Explanation:

Transform the left side of the expression:
$L S = {\sec}^{4} x - {\tan}^{4} x = \left({\sec}^{2} x - {\tan}^{2} x\right) \left({\sec}^{2} x + {\tan}^{2} x\right)$.
Since the first factor,
$\left({\sec}^{2} x - {\tan}^{2} x\right) = \left(\frac{1}{{\cos}^{2} x} - \frac{{\sin}^{2} x}{{\cos}^{2} x}\right) =$
$= \frac{1 - {\sin}^{2} x}{{\cos}^{2} x} = \frac{{\cos}^{2} x}{{\cos}^{2} x} = 1$
There for, the left side becomes;
$L S = \left({\sec}^{2} x + {\tan}^{2} x\right)$, and it equals the right side.