Trigonometric Identities Question 66
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 }= $
[Roorkee 1973]
Options:
A) $ \tan 3\theta $
B) $ \cot 3\theta $
C) $ \tan 6\theta $
D) $ \cot 6\theta $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \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 )} $ $ =\frac{2,\sin ,6\theta ,\cos ,3\theta +2,\sin ,6\theta ,\cos ,\theta }{2,\cos ,6\theta ,\cos ,3\theta +2,\cos ,6\theta ,\cos ,\theta } $ $ =\frac{2,\sin ,6\theta ,(\cos ,3\theta +\cos \theta )}{2,\cos ,6\theta ,(\cos ,3\theta +\cos \theta )}=\tan ,6\theta $ .
BETA
