Question

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

Answer 1

To prove:- $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$

Proof:-

Taking L.H.S.-

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

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

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

$=\frac{\sin A}{\cos A}$

$=\tan A$

$=$ R.H.S.

Hence proved.