Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}}$ = tan θ

To prove:

$\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}}$ = tan θ

Solving L.H.S. of the above equation :

$\Rightarrow \dfrac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 (1 - cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 + 2cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{tan θ}\text{ (2 cos}^2 θ - 1)}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}}$ = tan θ.