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Answer 1

To Prove :

$\dagger\ \; \sf \red{\dfrac{sin\ \theta - 2\ sin^3\theta}{2\ cos^3\theta - cos\ \theta}= tan\ \theta}$

Proof :

LHS

$\longrightarrow \sf \dfrac{sin\ \theta - 2\ sin^3 \theta }{2\ cos^3 \theta - cos\ \theta}$

$\longrightarrow \sf \dfrac{sin\ \theta(1 - 2\ sin^2 \theta )}{cos\ \theta (2\ cos^2 \theta - 1)}$

$\bullet\ \; \sf \orange{2\ cos^2x=1+cos\ 2x}$

$\bullet\ \; \sf \green{2\ sin^2x=1-cos\ 2x}$

$\longrightarrow \sf \dfrac{sin\ \theta(cos\ 2\theta )}{cos\ \theta (cos\ 2\theta)}$

$\longrightarrow \sf \dfrac{sin\ \theta}{cos\ \theta}$

$\longrightarrow\ \sf tan\ \theta$

RHS

Answer 2

Given : $\sf {\bf{ \dfrac{\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } = \tan \theta }}$

To Prove : $\sf {\bf{ \dfrac{\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } = \tan \theta }}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀ $\sf {\bf{ \star \dfrac{\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } = \tan \theta }}$

Here ,

⠀⠀⠀⠀⠀⠀ $\sf {\bf{ \implies L.H.S \:=\:\dfrac{\:\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } }}$

⠀⠀⠀⠀⠀⠀ $\sf {\bf{ \implies R.H.S \:=\:\tan \theta }}$

⠀⠀⠀⠀⠀$\sf{\underline {\bf{ \star { Now \: By \:Solving \;L.H.S \: \::}}}}$

⠀⠀⠀ $\sf { \implies{ L.H.S\:=\:\dfrac{ \:\:\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } }}$

⠀⠀⠀ $\sf { \implies \dfrac{\:\sin \theta - 2 \sin^{3} \theta }{2\cos^{3} \theta - \cos\theta } }$

By taking $\sin \theta$ as common in numerator :

⠀⠀⠀ $\sf { \implies {\dfrac{\:\purple {\sin \theta - 2 \sin^{3} \theta} }{2\cos^{3} \theta - \cos\theta } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\purple {\sin \theta(1 - 2 \sin^{2} \theta)} }{2\cos^{3} \theta - \cos\theta } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 \sin^{2} \theta) }{2\cos^{3} \theta - \cos\theta } }}$

By taking $\cos \theta$ as common in denominator :

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 \sin^{2} \theta) }{\purple {2\cos^{3} \theta - \cos\theta} } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 \sin^{2} \theta) }{\purple {\cos \theta (2\cos^{2}\theta - 1)} } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 \sin^{2} \theta) } {\cos \theta (2\cos^{2}\theta - 1) } }}$

As , We know that ,

$1 - \cos ^{2} \theta =\sin^{2} \theta$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 \sin^{2} \theta) } {\cos \theta (2\cos^{2}\theta - 1) } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2(1 - \cos^{2} \theta)) } {\cos \theta (2\cos^{2}\theta - 1) } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta(1 - 2 + 2 \cos^{2} \theta) } {\cos \theta (2\cos^{2}\theta - 1) } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta( 2\cos^{2} \theta- 1) } {\cos \theta (2\cos^{2}\theta - 1) } }}$

⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta\cancel{( 2\cos^{2} \theta- 1)} } {\cos \theta \cancel{(2\cos^{2}\theta - 1) }} }}$

⠀⠀⠀⠀ $\sf { \implies {\dfrac{\:\sin \theta } {\cos \theta } }}$

As , We know that ,

$\dfrac{\sin\theta}{\cos\theta} = \tan \theta$

⠀⠀⠀⠀ $\underline {\boxed{\pink{ \sf {L.H.S = \tan \theta }}}}$

Therefore,

⠀⠀⠀⠀ $\sf { \implies {L.H.S = R.H.S }}$

⠀⠀⠀⠀ $\dag\:\:\sf { \bf {Hence, \:Proved \:! }}$

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