## Precalculus (6th Edition) Blitzer

The above statement can be explained by using the trigonometric identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$. After putting the value of $x={{15}^{0}}$ and rearranging the equation, we get: ${{\tan }^{2}}{{15}^{0}}$ $-{{\sec }^{2}}{{15}^{0}}$ is $-1$.