Question

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

Answer 1

L.H.S $=\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-cosA}$

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

$=\frac{\tan A[1-2(1-{\cos }^{2}A\left)\right]}{2{\cos }^{2}A-1}[\because {\sin }^{2}A+{\cos }^{2}A=1]$

$=\frac{\tan A[1-2+2{\cos }^{2}A]}{2{\cos }^{2}A-1}$

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

$=\tan A$

$=R.H.S$

$\therefore \frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$.

Hence Proved.