Question

Prove that 2cos^3Φ-cosΦ/sinΦ-2sin^3Φ=cotΦ

Answer 1

$LHS = \frac{ 2cos^{3}Φ-cosΦ}{sinΦ-2sin^{3}Φ}$

$= \frac{ cosΦ( 2cos^{2}Φ - 1)}{sinΦ(1- 2sin^{2}Φ )}$

$= cotΦ\big(\frac{ 2(1-sin^{2}Φ )-1}{(1- 2sin^{2}Φ)}\big)$

$= cotΦ\big(\frac{ 2-2sin^{2}Φ -1}{(1- 2sin^{2}Φ)}\big)$

$=cotΦ\big(\frac{ 1-2sin^{2}Φ }{1- 2sin^{2}Φ}\big)\\= cotΦ\\= RHS$

$Hence \:proved .$

•••♪

Answer 2

Solution

We Are Given,

$\sf \: \dfrac{2 {cos}^{3} \alpha - cos \: \alpha }{ sin \: \alpha \: - 2 {sin}^{3} \alpha } = cot \: \alpha$

LHS

$\sf \: \dfrac{2 {cos}^{3} \alpha - cos \: \alpha }{ sin \: \alpha \: - 2 {sin}^{3} \alpha } \\ \\ \dashrightarrow \: \sf \: \dfrac{cos \: \alpha (2 {cos}^{2} \alpha - 1) }{sin \: \alpha (1 - 2sin {}^{2} \alpha ) } \\ \\ \dashrightarrow \: \sf \: \dfrac{cos \: \alpha }{sin \: \alpha } \times \cancel{\dfrac{cos2 \alpha }{cos2 \alpha } } \\ \\ \dashrightarrow \: \sf \: \dfrac{cos \: \alpha }{sin \: \alpha } \\ \\ \large{\dashrightarrow \boxed { \boxed{\sf \: cot \: \alpha }}}$

NOTE

$\star \ \boxed{\boxed{\sf \: cos2 \alpha = {cos}^{2} \alpha - {sin}^{2} \alpha }}$

$\sf Now, \\ \longrightarrow \: \sf \: cos2 \alpha = (1 - sin {}^{2} \alpha ) - sin {}^{2} \alpha \\ \\ \longrightarrow \: \boxed{ \boxed{ \: \sf \: cos2 \alpha = 1 - 2 {sin}^{2} \alpha }}$

Similarly,

$\sf \: cos \alpha = {cos}^{2} \alpha - (1 - cos {}^{2} \alpha ) \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: cos2 \alpha = 2 {cos}^{2} \alpha - 1}}$

Hence,ProveD