Question

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

• A. $\tan 3\theta$
• B. $\cot 3\theta$
• C. $\tan 6\theta$
• D. $\cot 6\theta$

Answer 1

The correct option is C $\tan 6\theta$

$a=\frac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }=\frac{(\sin 3\theta +\sin 9\theta )+(\sin 5\theta +\sin 7\theta )}{(\cos 3\theta +\cos 9\theta )+(\cos 5\theta +\cos 7\theta )}$
Using identities $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\&\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
$\Rightarrow a=\frac{2\sin \left(\frac{3\theta +9\theta }{2}\right)\cos \left(\frac{3\theta -9\theta }{2}\right)+2\sin \left(\frac{5\theta +7\theta }{2}\right)\cos \left(\frac{5\theta -7\theta }{2}\right)}{2\cos \left(\frac{3\theta +9\theta }{2}\right)\cos \left(\frac{3\theta -9\theta }{2}\right)+2\cos \left(\frac{5\theta +7\theta }{2}\right)\cos \left(\frac{5\theta -7\theta }{2}\right)}$
$\Rightarrow a=\frac{\sin 6\theta \cos 3\theta +\sin 6\theta \cos \theta }{\cos 6\theta \cos 3\theta +\cos 6\theta \cos \theta }=\tan 6\theta$
Ans: C