A) \[3\]
B) \[2\]
C) \[-1\]
D) \[0\]
E) \[1\]
Correct Answer: D
Solution :
Given, \[\sin \theta =3\sin (\theta +2\alpha )\] \[\Rightarrow \] \[\frac{\sin \theta }{\sin (\theta +2\alpha )}=\frac{3}{1}\] Use componendo and dividendo formula, \[=\frac{\sin \theta +\sin (\theta +2\alpha )}{\sin \theta -\sin (\theta +2\alpha )}=\frac{3+1}{3-1}\] \[=-\frac{2\sin (\theta +\alpha ).\cos \alpha }{2\cos (\theta +\alpha ).\sin \alpha }=2\] \[\Rightarrow \] \[-\tan (\theta +\alpha ).\cot \alpha =2\] \[\Rightarrow \] \[\tan (\theta +\alpha )=-2\tan \alpha \] \[\Rightarrow \] \[\tan (\theta +\alpha )=2\tan \alpha =0\]
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