Question

$\frac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }=\tan 6\theta$

Answer 1

$\frac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }$

$=\frac{2\sin 4\theta \cos \theta +2\sin 8\theta \cos \theta }{2\cos 4\theta \cos \theta +2\cos 8\theta +2\cos 8\theta \cos \theta }$

$=\frac{2\cos \theta (\sin 4\theta +\sin 8\theta )}{2\cos \theta (\cos 4\theta +\cos 8\theta )}$

$=\frac{2\sin 6\theta \cos 2\theta }{2\cos 6\theta \cos 2\theta }$

$=\tan 6\theta$