Question

$\frac{ \sin \: \theta - 2 \: { \sin \: }^{3} \theta }{2 \: { \cos }^{3} \theta \: - \: \cos \theta } = \tan \theta$
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Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

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

${\bf{\blue{\underline{Take\:L.H.S:}}}}$

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

${\sf{ Take \sin \theta \: and \: cos \theta \: common}}$

$={\sf{ \frac{sin \theta(1 - 2 {sin}^{2} \theta)}{cos \theta(2 {cos}^{2} \theta - 1) } }}$

$\boxed{\bf{\orange{\underline{ {sin}^{2} \theta = 1 - {cos}^{2} \theta }}}}$

$={\sf{ \frac{sin \theta \big[1 - 2(1 - {cos}^{2} \theta)\big] \: }{cos \theta(2 {cos}^{2} \theta - 1) } }}$

$={\sf{ \frac{sin \theta \big[1 - 2 + 2{cos}^{2} \theta\big] \: }{cos \theta(2 {cos}^{2} \theta - 1) } }}$

$={\sf{ \frac{sin \theta [ - 1 + 2{cos}^{2} \theta] \: }{cos \theta(2 {cos}^{2} \theta - 1) } }}$

$={\sf{ \frac{sin \theta ( 2{cos}^{2} \theta - 1) \: }{cos \theta(2 {cos}^{2} \theta - 1) } }}$

$={\sf{ \frac{sin \theta \cancel{( 2{cos}^{2} \theta - 1) }\: }{cos \theta \cancel{(2 {cos}^{2} \theta - 1)} } }}$

$={\sf{ \frac{sin \theta \: }{cos \theta} }}$

$={\sf{ tan \theta }}$

Hence L.H.S = R.HS

Answer 2

Answer :

Solving LHS,

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

Taking common,

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

$\leadsto\sf{=\dfrac{sin\theta}{cos\theta}\times\dfrac{1 - 2sin^{2}\theta}{2cos^{2}\theta - 1}}$

We know that, sin∅/cos∅ = tan∅.

Hence,

$\leadsto\sf{=tan\theta\times\dfrac{1 - 2sin^{2}\theta}{2cos^{2}\theta - 1}}$

We know that, sin²∅+cos²∅=1.

Hence, using it :

$\leadsto\sf{=tan\theta\times\dfrac{1 - 2(1 - cos^{2}\theta)}{2cos^{2}\theta - 1}}$

$\leadsto\sf{=tan\theta\times\dfrac{1 - 2 + 2cos^{2}\theta}{2cos^{2}\theta - 1}}$

$\leadsto\sf{=tan\theta\times\dfrac{- 1 + 2cos^{2}\theta}{2cos^{2}\theta - 1}}$

$\leadsto\sf{=tan\theta\times\dfrac{- 1 + 2cos^{2}\theta}{- 1 + 2cos^{2}\theta}}$

$\leadsto\underline{\bf{=tan\theta}}$

Hence, Proved.