Answer
$(\sec\theta-1)(\sec\theta+1)=\tan^{2}\theta$
Work Step by Step
$(\sec\theta-1)(\sec\theta+1)=\tan^{2}\theta$
The left side of the equation is the factored form of a difference of squares. The formula for factoring a difference of squares is $A^{2}-B^{2}=(A-B)(A+B)$. For this expression, $A=\sec\theta$ and $B=1$
Substitute the known values into $A^{2}-B^{2}$:
$(\sec\theta)^{2}-(1)^{2}=\tan^{2}\theta$
$\sec^{2}\theta-1=\tan^{2}\theta$
Since $\sec^{2}\theta-1=\tan^{2}\theta$, the identity is proved
