Question

Prove:
$\frac{tan5A-tan3A}{tan5A+tan3A}=\frac{sin2A}{sin8A}$

Answer 1

L.H.S

$=\frac{\tan 5A-\tan 3A}{\tan 5A+\tan 3A}$

$=\frac{\frac{\sin 5A}{\cos 5A}-\frac{\sin 3A}{\cos 3A}}{\frac{\sin 5A}{\cos 5A}+\frac{\sin 3A}{\cos 3A}}$

$=\frac{\sin 5A\cos 3A-\sin 3A\cos 5A}{\sin 5Acos3A+\sin 3A\cos 5A}$

We know that

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

Therefore,

$=\frac{\sin (5A-3A)}{\sin (5A+3A)}$

$=\frac{\sin 2A}{\sin 8A}$

R.H.S

Hence, proved.