Question

Prove that $\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2cosecA$

Answer 1

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

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

$=\frac{\tan A-\tan A\sec A-\tan A-\tan A\sec A}{1-{\sec }^{2}A}$

$=\frac{-2\tan A\sec A}{-{\tan }^{2}A}$ since ${\sec }^{2}A-1={\tan }^{2}A$

$=\frac{2\sec A}{\tan A}$

$=\frac{2\times \frac{1}{\cos A}}{\frac{\sin A}{\cos A}}$

$=2\csc A=$R.H.S

Hence proved.