Question
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $( cosec\; \theta - \cot \theta ) ^ { 2 } = \frac { 1 - \cos \theta } { 1 + \cos \theta }$
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Answer
$L H S = ( cosec\; \theta - \cot \theta ) ^ { 2 }$
$= \left( \frac { 1 } { \sin \theta } - \frac { \cos \theta } { \sin \theta } \right) ^ { 2 } = \left( \frac { 1 - \cos \theta } { \sin \theta } \right) ^ { 2 } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { \sin ^ { 2 } \theta }$
$= \frac { ( 1 - \cos \theta ) ^ { 2 } } { 1 - \cos ^ { 2 } \theta } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { ( 1 - \cos \theta ) ( 1 + \cos \theta ) }$
$= \frac { 1 - \cos \theta } { 1 + \cos \theta } = R H S$
$= \left( \frac { 1 } { \sin \theta } - \frac { \cos \theta } { \sin \theta } \right) ^ { 2 } = \left( \frac { 1 - \cos \theta } { \sin \theta } \right) ^ { 2 } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { \sin ^ { 2 } \theta }$
$= \frac { ( 1 - \cos \theta ) ^ { 2 } } { 1 - \cos ^ { 2 } \theta } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { ( 1 - \cos \theta ) ( 1 + \cos \theta ) }$
$= \frac { 1 - \cos \theta } { 1 + \cos \theta } = R H S$
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