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Prove that : $\frac{1 + \cot^2 A}{1 + \tan^2 A} = (\frac{1 - \cot A}{1 - \tan A})^2$
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Solution: LHS $= \frac{1 + \frac{\cos^2 A}{\sin^2 A}}{1 + \frac{\sin^2 A}{\cos^2 A}} = \frac{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}}{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}} = \frac{\frac{1}{\sin^2 A}}{\frac{1}{\cos^2 A}} = \frac{\cos^2 A}{\sin^2 A}$
$= \frac{\cos^2 A}{\sin^2 A} (\frac{\sin A - \cos A}{\cos A - \sin A})^2 = (\frac{\sin A - \cos A}{\sin A} \cdot \frac{\cos A}{\cos A - \sin A})^2 = (\frac{1 - \cot A}{1 - \tan A})^2 = RHS$
$= \frac{\cos^2 A}{\sin^2 A} (\frac{\sin A - \cos A}{\cos A - \sin A})^2 = (\frac{\sin A - \cos A}{\sin A} \cdot \frac{\cos A}{\cos A - \sin A})^2 = (\frac{1 - \cot A}{1 - \tan A})^2 = RHS$