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Prove that : $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \operatorname{cosec} A$
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LHS $$\begin{aligned}& = \frac{\tan A(1 - \sec A) - \tan A(1 + \sec A)}{1 - \sec^2 A} \\ & = \frac{-2 \tan A \sec A}{-\tan^2 A} \\ & = 2 \times \frac{\cos A}{\sin A} \times \frac{1}{\cos A} \\ & = 2 \operatorname{cosec} A = \text{RHS}\end{aligned}$$