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

Dec 13, 2015

The Pythagorean identity

${\sin}^{2} x + {\cos}^{2} x = 1$

can be divided by ${\cos}^{2} x$ to see that

${\tan}^{2} x + 1 = {\sec}^{2} x$

Then, $1$ can be subtracted from either side to see that

${\tan}^{2} x = {\sec}^{2} x - 1$

So, the identity on the left can be rewritten as

$\frac{{\sec}^{2} x - 1}{\sec x - 1}$

Now, see that ${\sec}^{2} x - 1$ is a difference of squares which can be factored into

$\left(\sec x + 1\right) \left(\sec x - 1\right)$

Substitute this into the expression to get

$\frac{\left(\sec x + 1\right) \left(\sec x - 1\right)}{\sec x - 1}$

Which simplifies to be

$\left(\sec x + 1\right)$