# How do you prove (sec+tan)(sec-tan)=1?

Jul 18, 2016

We will be using the following:

• $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

• ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$

With those:

$\left(\sec \left(\theta\right) + \tan \left(\theta\right)\right) \left(\sec \left(\theta\right) - \tan \left(\theta\right)\right)$

$= {\sec}^{2} \left(\theta\right) - {\tan}^{2} \left(\theta\right)$

$= \left({\tan}^{2} \left(\theta\right) + 1\right) - {\tan}^{2} \left(\theta\right)$

$= 1$