A) \[\frac{m+n}{m-n}\]
B) \[\frac{m-n}{m+n}\]
C) \[\frac{1-n}{1+n}\]
D) \[\frac{1+n}{1-n}\]
Correct Answer: A
Solution :
Given,\[m\text{ }sin\text{ }\theta =n\text{ }sin(\theta +2\alpha )\] \[\frac{m}{n}=\frac{\sin (\theta +2\alpha )}{\sin \theta }\] Applying componendo-dividendo rule \[\frac{m+n}{m-n}=\frac{\sin (\theta +2\alpha )+\sin \theta }{\sin (\theta +2\alpha )-\sin \theta }\] \[\Rightarrow \] \[\frac{m+n}{m-n}=\frac{2\sin (\theta +\alpha )\cos \alpha }{2\sin (\theta +\alpha )\sin \alpha }\] \[\Rightarrow \] \[\frac{\sin (\theta +\alpha )\cos \alpha }{\cos (\theta +\alpha )\sin \alpha }=\frac{m+n}{m-n}\] \[\Rightarrow \] \[\tan (\theta +\alpha ).\cot \alpha =\frac{m+n}{m-n}\]
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