Question

Prove that $\frac{\mathrm{tanA}}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\text{cosec}A$

Answer 1

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

L.H.S

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

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

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

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

$=\frac{2\cdot 1/cosA}{\sin A/cosA}$

$=\frac{2}{\sin A}$

$=2\cdot cosecA$

$=$R.H.S

$\therefore$ L.H.S$=$R.H.S.