Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:$\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}} = \text{tan θ}$

Answer 1

$\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}} = \text{tan θ}$

L.H.S =$\frac{\text{(sinθ - 2sin³θ)}}{\text{(2cos³θ - cosθ)}}$

$⇒ \frac{\text{sinθ (1 - 2sin²θ)}}{\text{cosθ {2(1 - sin²θ) - 1}}}$

$= \frac{\text{sinθ (1 - 2sin²θ)}}{\text{cosθ(2 - 2sin²θ - 1)}}$

$=\frac{\text{ sinθ (1 - 2sin²θ)}}{\text{cosθ(1 - 2sin²θ)}}$

$=\frac{ \text{sinθ}}{\text{cosθ}}$

$=\text{ tanθ}$
= R.H.S