185
Prove that : $( \frac{1}{\cos A} - \cos A ) ( \frac{1}{\sin A} - \sin A ) = \frac{1}{\tan A + \cot A}$
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Sol. LHS $= (\frac{1 - \cos^2 A}{\cos A}) (\frac{1 - \sin^2 A}{\sin A})$
$= (\frac{\sin^2 A}{\cos A}) (\frac{\cos^2 A}{\sin A})$
$= \sin A \cos A$
RHS $= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$
$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{1}$
$= \sin A \cos A$
$\therefore$ LHS $=$ RHS
$= (\frac{\sin^2 A}{\cos A}) (\frac{\cos^2 A}{\sin A})$
$= \sin A \cos A$
RHS $= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$
$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{1}$
$= \sin A \cos A$
$\therefore$ LHS $=$ RHS