Question

Solve:

$\frac{\tan A}{1+\sec A}-\frac{\mathrm{tanA}}{1-\sec A}=2\csc A$

Answer 1

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

LHS:- $\tan A[\frac{1}{1+\sec A}-\frac{1}{1-\sec A}]$

$\Rightarrow \tan A\left[\frac{1-\sec A-1-\sec A}{1-{\sec }^{2}A}\right]$

$\Rightarrow \frac{\sin A}{\cos A}\left[\frac{-\frac{2}{\cos A}}{{\cos }^{2}A-1}\right]\times {\cos }^{2}A$

$\Rightarrow \frac{-2\sin A}{{\cos }^{2}A-1}$

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

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

$\Rightarrow 2cosecA=$RHS

$\Rightarrow$ LHS$=$RHS

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