Question

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

Answer 1

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

$L.H.S.$ $=\frac{(\tan A+\sec A)-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}$

$=\frac{(\tan A+\sec A)[1-(\sec A-\tan A)]}{\tan A-\sec A+1}$ $=\tan A+\sec A=\frac{\sin A+1}{\cos A}$.