Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$

(A) To prove: $(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
$L.H.S. = (\operatorname{cosec} A - \sin A)(\sec A - \cos A)$
$= (\frac{1}{\sin A} - \sin A)(\frac{1}{\cos A} - \cos A)$
$= (\frac{1 - \sin^2 A}{\sin A})(\frac{1 - \cos^2 A}{\cos A})$
$= (\frac{\cos^2 A}{\sin A})(\frac{\sin^2 A}{\cos A})$
$= \sin A \cos A$
$R.H.S. = \frac{1}{\tan A + \cot A}$
$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$
$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
$= \frac{\sin A \cos A}{1} = \sin A \cos A$
Since $L.H.S. = R.H.S.$,the identity is proved.