Trigonometric Identities Question 83
Question: If $ m\tan (\theta -30{}^\circ )=n\tan (\theta +120{}^\circ ), $ then $ \frac{m+n}{m-n}= $
[IIT 1966]
Options:
A) $ 2,\cos ,2\theta $
B) $ \cos 2\theta $
C) $ 2,\sin ,2\theta $
D) $ \sin 2\theta $
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Answer:
Correct Answer: A
Solution:
$ \frac{m}{n}=\frac{\tan ,(120^{o}+\theta )}{\tan ,(\theta -30^{o})} $
$ \Rightarrow \frac{m+n}{m-n}=\frac{\tan ,(\theta +120^{o})+\tan ,(\theta -30^{o})}{\tan ,(\theta +120^{o})-\tan ,(\theta -30^{o})} $ (By componendo and dividendo) $ =\frac{\sin (\theta +120^{o})\cos (\theta -30^{o})+\cos (\theta +120^{o})\sin (\theta -30^{o})}{\sin (\theta +120^{o})\cos (\theta -30^{o})-\cos (\theta +120^{o})\sin (\theta -30^{o})} $ $ =\frac{\sin ,(2\theta +90^{o})}{\sin ,(150^{o})}=\frac{\cos ,2\theta }{1/2}=2,\cos ,2\theta $ .
BETA
