Question
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. (cosec A - sin A) (sec A - cos A) = $\frac { 1 } { \tan A + \cot A }$
[Hint: Simplify LHS and RHS separately]

Answer

LHS
$= ( \csc A - \sin A ) ( \sec A - \cos A )$
$= \left( \frac { 1 } { \sin A } - \sin A \right) \left( \frac { 1 } { \cos A } - \cos A \right) = \frac { 1 - \sin ^ { 2 } A } { \sin A } \frac { 1 - \cos ^ { 2 } A } { \cos A }$
$= \frac { \cos ^ { 2 } A } { \sin A } \frac { \sin ^ { 2 } A } { \cos A } , \ldots \ldots \ \because \sin ^ { 2 } A + \cos ^ { 2 } A = 1 = \frac { \frac { \sin A \cos A } { \sin A \cos A } } { \frac { \sin ^ { 2 } A } { \sin A \cos A } + \frac { \cos ^ { 2 } A } { \sin A \cos A } }$
........ Dividing the numerator and denominator by sin A cos A
$= \frac { 1 } { \tan A + \cot A }$
= RHS

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