Answer
$sec^2~\alpha-1 = \frac{sec~2\alpha-1}{sec~2\alpha+1}$
Work Step by Step
We can verify that the equation is an identity:
$\frac{sec~2\alpha-1}{sec~2\alpha+1}$
$=\frac{\frac{1}{cos~2\alpha}-1}{\frac{1}{cos~2\alpha}+1}$
$=\frac{\frac{1-cos~2\alpha}{cos~2\alpha}}{\frac{1+cos~2\alpha}{cos~2\alpha}}$
$=\frac{1-cos~2\alpha}{1+cos~2\alpha}$
$=\frac{1-(2cos^2~\alpha-1)}{1+(2cos^2~\alpha-1)}$
$= \frac{1-2cos^2~\alpha+1}{1+2cos^2~\alpha-1}$
$= \frac{2-2cos^2~\alpha}{2cos^2~\alpha}$
$= \frac{2}{2cos^2~\alpha}- \frac{2cos^2~\alpha}{2cos^2~\alpha}$
$= \frac{1}{cos^2~\alpha}-1$
$= sec^2~\alpha-1$
