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Prove that $\frac{\cos \theta - 2\cos^3 \theta}{\sin \theta - 2\sin^3 \theta} + \cot \theta = 0$.
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$LHS = \frac{\cos \theta - 2\cos^3 \theta}{\sin \theta - 2\sin^3 \theta} + \cot \theta = \frac{\cos \theta(1 - 2\cos^2 \theta)}{\sin \theta(1 - 2\sin^2 \theta)} + \cot \theta$ ($\frac{1}{2}$ mark).
$= \frac{\cos \theta}{\sin \theta} [\frac{\sin^2 \theta + \cos^2 \theta - 2\cos^2 \theta}{\sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta}] + \cot \theta$ (1 mark).
$= \frac{\cot \theta(\sin^2 \theta - \cos^2 \theta)}{(\cos^2 \theta - \sin^2 \theta)} + \cot \theta$ (1 mark).
$$\begin{aligned}& = -\cot \theta + \cot \theta \\ & = 0 = RHS\end{aligned}$$ ($\frac{1}{2}$ mark).
$= \frac{\cos \theta}{\sin \theta} [\frac{\sin^2 \theta + \cos^2 \theta - 2\cos^2 \theta}{\sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta}] + \cot \theta$ (1 mark).
$= \frac{\cot \theta(\sin^2 \theta - \cos^2 \theta)}{(\cos^2 \theta - \sin^2 \theta)} + \cot \theta$ (1 mark).
$$\begin{aligned}& = -\cot \theta + \cot \theta \\ & = 0 = RHS\end{aligned}$$ ($\frac{1}{2}$ mark).