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Prove that $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\cot A-\tan A}$
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LHS $= (\frac{1}{\sin A} - \sin A) (\frac{1}{\cos A} - \cos A)$
$= (\frac{1 - \sin^2 A}{\sin A}) (\frac{1 - \cos^2 A}{\cos A})$
$= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A}$
$= \sin A \cos A$
RHS $= \frac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}$
$= \frac{1}{\frac{\cos^2 A + \sin^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{1}$
$= \sin A \cos A = \text{LHS}$
$= (\frac{1 - \sin^2 A}{\sin A}) (\frac{1 - \cos^2 A}{\cos A})$
$= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A}$
$= \sin A \cos A$
RHS $= \frac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}$
$= \frac{1}{\frac{\cos^2 A + \sin^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{1}$
$= \sin A \cos A = \text{LHS}$