Prove that sin A - 2 sin^3 A / 2cos^3 A

Q) Prove that $\frac{\sin A - 2\sin^3 A}{2\cos^3 A-\cos A} = \tan A$

Ans:

Let’s start from LHS

LHS = $\frac{\sin A - 2\sin^3A}{2\cos^3A-\cos A}$

= $\frac{\sin A (1- 2\sin^2 A)}{\cos A(2\cos^2A-1)}$

= $(\frac{\sin A}{\cos A})(\frac{1 - 2\sin^2A}{2\cos^2A-1})$

We know that sin²A + cos²A = 1

$\therefore$ LHS = $(\tan A)(\frac{(\sin^2A + \cos^2A) - 2\sin^2A}{2\cos^2A -(\sin^2A + \cos^2A)})$

= $(\tan A)(\frac{\cos^2A - \sin^2A}{\cos^2A - \sin^2A})$

=tan A …….. RHS

Hence proved!

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