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Prove that :
$\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \sec A \cosec A$
$\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \sec A \cosec A$
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$$\begin{aligned}& LHS = \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} \\ & = \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} \\ & = \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)} \\ & = \frac{1}{(\sin A - \cos A)} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right] \\ & = \frac{1}{(\sin A - \cos A)} \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{\sin A \cos A} \\ & = \frac{1 + \sin A \cos A}{\sin A \cos A} \\ & = \frac{1}{\sin A \cos A} + 1 \\ & = 1 + \sec A \cosec A = RHS\end{aligned}$$