A) \[\frac{\tan \beta }{\tan (\alpha +\beta )}\]
B) \[\frac{\cot \beta }{\cot (\alpha -\beta )}\]
C) \[\frac{-\cot \beta }{\cot (\alpha +\beta )}\]
D) \[\frac{\cot \beta }{\cot (\alpha +\beta )}\]
Correct Answer: C
Solution :
We have \[b\,\sin \,\alpha =a\,\sin \,(\alpha +2\beta )\,\Rightarrow \,\frac{a}{b}=\frac{\sin \,\alpha }{\sin \,(\alpha +2\beta )}\] \[\Rightarrow \,\,\frac{a+b}{a-b}=\frac{\sin \,\alpha +\sin \,(\alpha +2\beta )}{\sin \,\alpha -\sin \,(\alpha +2\beta )}=\frac{2\,\sin \,(\alpha +\beta )\,\cos \,\beta }{-2\,\cos \,(\alpha +\beta )\,\sin \,\beta }\] \[=-\tan \,(\alpha +\beta )\,\cot \,\beta =-\frac{\cot \beta }{\cot \,(\alpha +\beta )}\].
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