Question
Prove.
$\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}=\tan A$
$\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}=\tan A$
✓
Answer
$\text { LHS }=\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A} $
$ =\frac{\sin A\left(1-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-1\right)} $
$ =\frac{\sin A\left(\sin ^2 A+\cos ^2 A-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-\sin ^2 A-\cos ^2 A\right)} $
$=\frac{\sin A\left(\cos ^2 A-\sin ^2 A\right)}{\cos A\left(\cos ^2 A-\sin ^2 A\right)} $
$=\frac{\sin A}{\cos A} $
$=\text { tan A = RHS }$
$ =\frac{\sin A\left(1-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-1\right)} $
$ =\frac{\sin A\left(\sin ^2 A+\cos ^2 A-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-\sin ^2 A-\cos ^2 A\right)} $
$=\frac{\sin A\left(\cos ^2 A-\sin ^2 A\right)}{\cos A\left(\cos ^2 A-\sin ^2 A\right)} $
$=\frac{\sin A}{\cos A} $
$=\text { tan A = RHS }$
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