# How do you simplify the expression sec^2theta-tan^2theta+cot^2theta?

Sep 21, 2016

This expression simplifies to ${\csc}^{2} \theta$.

#### Explanation:

Apply the following identities:

$\sec \theta = \frac{1}{\cos} \theta$
$\tan \theta = \sin \frac{\theta}{\cos} \theta$
$\cot \theta = \cos \frac{\theta}{\sin} \theta$

$= \frac{1}{\cos} ^ 2 \theta - {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

$= \frac{1 - {\sin}^{2} \theta}{{\cos}^{2} \theta} + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

Apply the pythagorean identity ${\sin}^{2} x + {\cos}^{2} x = 1 \to 1 - {\sin}^{2} x = {\cos}^{2} x$

$= {\cos}^{2} \frac{\theta}{\cos} ^ 2 \theta + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

$= 1 + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta$

Apply the identity ${\sin}^{2} \beta + {\cos}^{2} \beta = 1$:

$= \frac{1}{\sin} ^ 2 \theta$

Apply the identity $\frac{1}{\sin} \alpha = \csc \alpha$.

$= {\csc}^{2} \theta$

Hopefully this helps!