Question

$\frac{tanA+secA-1}{tanA-secA+1}=\frac{1+sinA}{cosA}$

Answer 1

$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$

Taking L.H.S.-

$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$

$=\frac{(\tan A+\sec A)-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}[\because 1+{\tan }^{2}A={\sec }^{2}A]$

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

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

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

$=\tan A+\sec A$

$=\frac{\sin A}{\cos A}+\frac{1}{\cos A}$

$=\frac{1+\sin A}{\cos A}$

$=$ R.H.S.

Hence proved.