Question

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

Answer 1

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

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

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

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

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

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

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

$=\frac{\cos A}{1-\sin A}\times \frac{1+sinA}{1+\sin A}$

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

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

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