Question

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

Answer 1

$$\begin{array}{c}\frac{\tan A+\sec A-1}{\tan A+\sec A-1}=\frac{1+\sin A}{\cos A}\\ \frac{\tan A+\sec A-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}\\ \frac{(\tan A+\sec A)(\sec A-\tan A)(\sec A+\tan A}{\tan A-seA+1}\\ \frac{(\tan A+\sec A)(1+\tan A-\sec A)}{(\tan A-\sec A+1)}\\ \frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{1+\sin A}{\cos A}Ans.\end{array}$$