Question

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

Answer 1

LHS

$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}\frac{\tan A+\sec A-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}\frac{\tan A+\sec A-(\tan A+\sec A)(\sec A-\tan A)}{\tan A-\sec A+1}\frac{\tan A+\sec A(1-\sec A+\tan A)}{\tan -\sec A+1}\tan A+\sec A\frac{\sin A}{\cos A}+\frac{1}{\cos A}\frac{1+\sin A}{\cos A}$

RHS