## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\sec }^{2}}\theta -1$ and using the identity ${{\sec }^{2}}\theta ={{\tan }^{2}}\theta +1$ we simplify as: \begin{align} & \left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\sec }^{2}}\theta -1 \\ & =1+{{\tan }^{2}}\theta -1 \\ & ={{\tan }^{2}}\theta \end{align} Hence, the expression on the left side is equal to the expression on the right side $\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\tan }^{2}}\theta$.