# How do you prove sec^4(x)-csc^2(x)=tan^2(x)-cot^2(x)?

Oct 25, 2015

See explanation.

#### Explanation:

Are you sure you didn't mean ${\sec}^{2} x - {\csc}^{2} x = {\tan}^{2} x - {\cot}^{2} x$? I don't think the data you provided is an identity. Anyway, if you meant ${\sec}^{2} x$, here it is:

$\left[1\right] \text{ } {\sec}^{2} x - {\csc}^{2} x$

Pythagorean Identity: ${\sec}^{2} \alpha = 1 + {\tan}^{2} \alpha$

$\left[2\right] \text{ } \equiv \left(1 + {\tan}^{2} x\right) - {\csc}^{2} x$

Pythagorean Identity: ${\csc}^{2} \alpha = 1 + {\cot}^{2} \alpha$

$\left[3\right] \text{ } \equiv \left(1 + {\tan}^{2} x\right) - \left(1 + {\cot}^{2} x\right)$

$\left[4\right] \text{ } = 1 + {\tan}^{2} x - 1 - {\cot}^{2} x$

$\left[5\right] \text{ } = {\tan}^{2} x - {\cot}^{2} x$

$\textcolor{b l u e}{\therefore {\sec}^{2} x - {\csc}^{2} x = {\tan}^{2} x - {\cot}^{2} x}$