Question

Prove that $(\sec A+\tan A-1)(\sec A-\tan A+1)-2\tan A=0$

Answer 1

$L.H.S.=[\sec A+(\tan A-1\left)\right]$
$[\sec A-(\tan A-1\left)\right]-2\tan A$
$={\sec }^{2}A-(\tan A-1{)}^2-2\tan A$
${\sec }^{2}A-{\tan }^{2}A-1+2\tan A-2\tan A$
$=1-1=0=R.H.S.$