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Q. If $\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}=\tan k \theta$ then find value of $k$.

Trigonometric Functions

Solution:

$\frac{\sin 9 \theta+\sin 7 \theta+\sin 5 \theta+\sin 3 \theta}{\cos 9 \theta+\cos 7 \theta+\cos 5 \theta+\cos 3 \theta}$
$=\frac{2 \sin 8 \theta \cos \theta+2 \sin 4 \theta \cos \theta}{2 \cos 8 \theta \cos \theta+2 \cos 4 \theta \cos \theta}$
$=\frac{2 \cos \theta(\sin 8 \theta+\sin 4 \theta)}{2 \cos \theta(\cos 8 \theta+\cos 4 \theta)}$
$=\frac{2 \sin 6 \theta \cos 2 \theta}{2 \cos 6 \theta \cos 2 \theta}$
$=\tan 6\, \theta$