# How do you evaluate sec^2(18) - tan^2(18)?

${\sec}^{2} \left(18\right) - {\tan}^{2} \left(18\right)$
$= \frac{1}{\cos} ^ 2 \left(18\right) - {\sin}^{2} \frac{18}{\cos} ^ 2 \left(18\right)$
$= \frac{1 - {\sin}^{2} \left(18\right)}{\cos} ^ 2 \left(18\right)$
$= {\cos}^{2} \frac{18}{\cos} ^ 2 \left(18\right)$
$= 1$