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Prove the following trigonometric identity :
$\frac{\cos A - 2 \cos^3 A}{2 \sin^3 A - \sin A} = \cot A$
$\frac{\cos A - 2 \cos^3 A}{2 \sin^3 A - \sin A} = \cot A$
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Solution: $LHS = \frac{\cos A (1 - 2 \cos^2 A)}{\sin A (2 \sin^2 A - 1)}$ [1/2 mark]
$= \frac{\cos A [1 - 2 (1 - \sin^2 A)]}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \frac{\cos A (- 1 + 2 \sin^2 A)}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \cot A$ [1/2 mark]
$= \frac{\cos A [1 - 2 (1 - \sin^2 A)]}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \frac{\cos A (- 1 + 2 \sin^2 A)}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \cot A$ [1/2 mark]