Problem Statement
Prove the identity: $\frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta$
Verified Solution & Marking Scheme
Factor common trigonometric terms
$\text{LHS} = \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}$
Express numerator in terms of cos²θ using sin²θ = 1 - cos²θ
$1 - 2\sin^2\theta = 1 - 2(1 - \cos^2\theta) = 1 - 2 + 2\cos^2\theta = 2\cos^2\theta - 1$
Substitute into the expression:
$\text{LHS} = \frac{\sin\theta(2\cos^2\theta - 1)}{\cos\theta(2\cos^2\theta - 1)}$
Cancel common binomial factor and conclude
$\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$
Hence proved.