Question

$\frac{tan5\theta +tan3\theta }{tan5\theta -tan3\theta }=4cos2\theta cos4\theta$

Answer 1

L.H.S$=\frac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }$

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

$=\frac{\frac{\sin 5\theta \cos 3\theta +cos5\theta \sin 3\theta }{cos5\theta \cos 3\theta }}{\frac{\sin 5\theta \cos 3\theta -cos5\theta \sin 3\theta }{cos5\theta \cos 3\theta }}$

$=\frac{\sin 5\theta \cos 3\theta +cos5\theta \sin 3\theta }{\sin 5\theta \cos 3\theta -cos5\theta \sin 3\theta }$

$=\frac{\sin (5\theta +3\theta )}{\sin (5\theta -3\theta )}$ using compound angle formula

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

$=\frac{2\sin 4\theta \cos 4\theta }{\sin 2\theta }$ using multiple angle formula

$=\frac{2\times 2\sin 2\theta \cos 2\theta \cos 4\theta }{\sin 2\theta }$ using multiple angle formula

$=4\cos 2\theta \cos 4\theta =$R.H.S

Hence proved.