Trigonometric Identities Question 72
Question: If $ b\sin \alpha =a\sin (\alpha +2\beta ), $ then $ \frac{a+b}{a-b}= $
Options:
A) $ \frac{\tan \beta }{\tan (\alpha +\beta )} $
B) $ \frac{\cot \beta }{\cot (\alpha -\beta )} $
C) $ \frac{-\cot \beta }{\cot (\alpha +\beta )} $
D) $ \frac{\cot \beta }{\cot (\alpha +\beta )} $
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Answer:
Correct Answer: C
Solution:
We have $ b,\sin ,\alpha =a,\sin ,(\alpha +2\beta ),\Rightarrow ,\frac{a}{b}=\frac{\sin ,\alpha }{\sin ,(\alpha +2\beta )} $
$ \Rightarrow \frac{a+b}{a-b}=\frac{\sin ,\alpha +\sin ,(\alpha +2\beta )}{\sin ,\alpha -\sin ,(\alpha +2\beta )}=\frac{2,\sin ,(\alpha +\beta ),\cos ,\beta }{-2,\cos ,(\alpha +\beta ),\sin ,\beta } $ $ =-\tan ,(\alpha +\beta ),\cot ,\beta =-\frac{\cot \beta }{\cot ,(\alpha +\beta )} $ .
BETA
