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Prove that : $\frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A} = 2 \cosec A$
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$LHS = \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A} = \frac{\frac{\sin A}{\cos A}}{1+\frac{1}{\cos A}} - \frac{\frac{\sin A}{\cos A}}{1-\frac{1}{\cos A}}$ (I) (1 Mark)
$= \frac{\sin A}{\cos A+1} - \frac{\sin A}{\cos A-1}$ (II) (1/2 Mark)
$= \sin A (\frac{-2}{\sin^2A})$ (III) (1 Mark)
$= \frac{2}{\sin A} = 2 \cosec A = RHS$ (IV) (1/2 Mark)
$= \frac{\sin A}{\cos A+1} - \frac{\sin A}{\cos A-1}$ (II) (1/2 Mark)
$= \sin A (\frac{-2}{\sin^2A})$ (III) (1 Mark)
$= \frac{2}{\sin A} = 2 \cosec A = RHS$ (IV) (1/2 Mark)