Answer
Work on the left side of the identity using $\sec{\theta}=\frac{1}{\cos\theta}$ and $1-\sin^\theta=\cos^2\theta$.
Refer to the step-by-step part below for the complete proof.
Work Step by Step
We have to show that:
$(\sec\theta+1)(\sec\theta-1)=\tan^2\theta$
By evaluating the left side we get:
$=\sec^2\theta+\sec\theta-\sec\theta-1\\
=\sec^2\theta-1$
Since $\sec\theta=\frac{1}{\cos\theta}$ and $1-\cos^2\theta=\sin^2\theta$, the expression above simplifies to:
$=\dfrac{1}{\cos^2\theta}-\dfrac{\cos^2\theta}{\cos^2\theta}\\
=\dfrac{1-\cos^2\theta}{\cos^2\theta}\\
=\dfrac{\sin^2\theta}{\cos^2\theta}\\
=\tan^2\theta$
Since LHS=RHS, the proof is complete
