Question

Prove that:
2 cos³∅ - cos∅ /sin∅ - 2 sin³∅ == cot∅​

Answer 1

Question:

$\sf{Prove\:that\:\dfrac{2cos^3\theta-cos\theta}{sin\theta-2sin^3\theta}=cot\theta}$

Identities Used:

$\sf{2cos^2\theta-1=cos2\theta}\\\\\sf{1-2sin^2\theta=cos2\theta}\\\\\sf{\dfrac{cos\theta}{sin\theta}=cot\theta}$

Solution:

$\sf{Taking\:L.H.S.}$

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

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

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

$\sf{\implies\:\dfrac{cos\theta.cos2\theta}{sin\theta.cos2\theta}}$

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

$\sf{\implies\:cot\theta}$

$\mathbb{HENCE\:PROVED}$

Answer 2

$\large \bigstar\boxed{\bf \red{SOLUTION}} \bigstar \\ \\ \bf On \:Solving \:LHS \\ \\ \Longrightarrow \bf \frac{2 { \cos }^{3} \theta - \cos\theta }{ \sin\theta - 2 { \sin }^{3} \theta } \\ \\ \Longrightarrow \bf \frac{ \cos \theta(2 { \cos }^{2} \theta - 1) }{ \sin \theta(1 - 2 { \sin }^{2} x) } \\ \\\Longrightarrow \bf \frac{ \cos \theta(2 { \cos }^{2} \theta - 1) }{ \sin \theta \{1 - 2 (1 - { \cos }^{2}\theta ) \}} \\ \\\Longrightarrow \bf \frac{ \cos \theta(2 { \cos }^{2} \theta - 1) }{ \sin \theta(1 - 2 + 2 { \cos }^{2}\theta) } \\ \\ \Longrightarrow \bf \frac{ {\cos \theta \: \cancel{(2 { \cos }^{2}\theta - 1)} }}{ \sin \theta \: \cancel{( - 1 + 2 { \cos }^{2} \theta) }} \\ \\ \Longrightarrow \bf \frac{ \cos \theta }{ \sin \theta } = \cot \theta \\ \\ \bf Which \: is \: equals \: to \: RHS \\ \\ \bf Hence \: Proved$