Question

plz answer fast
prove that ⤵⤵⤵⤵⤵⤵↪↪↪↪↪↪
$\sin(theta ) - 2 \sin^{3} theta \div 2 \ \cos^{3} theta - \cos(theta) = \tan(theta)$
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Answer 1

We have to prove that :

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

On taking LHS :

$\frac{ \sin\alpha - 2 { \sin}^{3} \alpha }{2 { \cos }^{3} \alpha - \cos \alpha } \\ \\ = > \frac{ \sin\alpha (1 - 2 { \sin}^{2} \alpha )}{ \cos \alpha (2 { \cos }^{2} \alpha - 1) } \\ \\ As \: we \: know \: that: \: 2 { \cos }^{2} \alpha - 1 = 1 - 2 { \sin }^{2} \alpha = \cos2 \alpha \\ \\ = > \frac{ \sin\alpha \cos2 \alpha }{ \cos\alpha \cos2 \alpha } \\ \\ = > \tan\alpha = RHS \\ \\ HENCE \: PROVED$