Prove the given identity, where the angles involved are acute ang…

Question

Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { \sin \theta - 2 \sin ^ { 3 } \theta } { 2 \cos ^ { 2 } \theta - \cos \theta } = \tan \theta$

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Answer

LHS
$= \frac { \sin \theta - 2 \sin ^ { 3 } \theta } { 2 \cos ^ { 2 } \theta - \cos \theta } = \frac { \sin \theta \left( 1 - 2 \sin ^ { 2 } \theta \right) } { \cos \theta \left( 2 \cos ^ { 2 } \theta - 1 \right) }$
$= \frac { \sin \theta \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta - 2 \sin ^ { 2 } \theta \right) } { \cos \theta \left( 2 \cos ^ { 2 } \theta - \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } \quad \because \cos ^ { 2 } \theta + \sin ^ { 2 } \theta = 1$
$= \frac { \sin \theta \left( \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } { \cos \theta \left( \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } = \tan \theta$
= RHS

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