Question

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

Answer 1

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

$=(\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1)(\frac{1}{\cos A}-\frac{\sin A}{\cos A}+1)$

$=\left(\frac{1+\sin A-\cos A}{\cos A}\right)\left(\frac{1-\sin A+\cos A}{\cos A}\right)$

$=\frac{1-\sin A+\cos A+\sin A-{\sin }^{2}A+\sin A\cos A-\cos A+\sin A\cos A-{\cos }^{2}A}{{\cos }^{2}A}$

$=\frac{2\sin A\cos A}{{\cos }^{2}A}$

$=2\tan A$

$=RHS$

Hence proved