# How do you prove tan^2(x) / (sec(x) - 1) = (sec(x) + 1)?

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