Prove the following trigonometric identities.

We have to prove $(\sec A+\tan A-1)(\sec A-\tan A+1)=2 \tan A$

We know that, $\sec ^{2} A-\tan ^{2} A=1$

So, we have

$(\sec A+\tan A-1)(\sec A-\tan A+1)=\{\sec A+(\tan A-1)\}\{\sec A-(\tan A-1)\}$

$=\sec ^{2} A-(\tan A-1)^{2}$

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

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

So, we have

$(\sec A+\tan A-1)(\sec A-\tan A+1)=1+2 \tan A-1$

$=2 \tan A$

Hence proved.