Question

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Answer 1

TO PROVE :-

$\frac{ \sin \theta - 2 { \sin }^{3} \theta }{2 { \cos }^{3} - \cos \theta } = \tan \theta$

PROOF :-

LHS=

$\frac{ \sin \theta - 2 { \sin }^{3} \theta }{2 { \cos }^{3} - \cos \theta } \\ \\ \\ = \frac{ \sin \theta(1 - 2 { \sin }^{2} \theta)}{ \cos \theta(2 { \cos}^{2} \theta - 1 )} \\ \\ \\ = \frac{ \sin \theta(1 - 2 { \sin }^{2} \theta)}{ \cos \theta(2 - 2 { \sin }^{2} \theta - 1 )} \\ \\ \\ = \frac{ \sin \theta(1 - 2 { \sin }^{2} \theta)}{ \cos \theta(1 - 2 { \sin }^{2} \theta )} \\ \\ \\ = \frac{ \sin \theta }{ \cos \theta} \\ \\ = \boxed{ \tan \theta }$

= RHS

IDENTITY USED :-

$\frac{ \sin \theta }{ \cos \theta} = tan \theta \\ \\ \\ { \sin \theta}^{2} + { \cos \theta}^{2} = 1$