Trigonometric Identities Question 13
Question: $ \frac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }= $
[Roorkee 1971]
Options:
A) $ \frac{1}{2}\tan \theta $
B) $ \frac{1}{2}\cot \theta $
C) $ \tan \theta $
D) $ \cot \theta $
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Answer:
Correct Answer: C
Solution:
$ \frac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta } $ $ =\frac{\sin \theta +2\sin \theta \cos \theta }{2{{\cos }^{2}}\theta +\cos \theta }=\frac{\sin \theta (1+2\cos \theta )}{\cos \theta (1+2\cos \theta )}=\tan \theta $ . Trick: Put $ \theta =30{}^\circ $ , since for $ \theta =30{}^\circ $ no option will give the common value.
BETA
