195
Prove that $\frac{\cos \theta - 2\cos^3 \theta}{\sin \theta - 2\sin^3 \theta} + \cot \theta = 0$.
Show SolutionHide Solution↓
$LHS = \frac{\cos \theta - 2\cos^3 \theta}{\sin \theta - 2\sin^3 \theta} + \cot \theta = \frac{\cos \theta(1 - 2\cos^2 \theta)}{\sin \theta(1 - 2\sin^2 \theta)} + \cot \theta$ ($\frac{1}{2}$ 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). $= -\cot \theta + \cot \theta = 0 = RHS$ ($\frac{1}{2}$ mark).