A) 3
B) 2
C) \[-1\]
D) 0
Correct Answer: D
Solution :
\[\sin \theta =3\sin (\theta +2\alpha )\Rightarrow \sin (\theta +\alpha -\alpha )=3\sin \] \[(\theta +\alpha +\alpha )\] \[\Rightarrow \,\,\sin (\theta +\alpha )\cos \alpha -\cos (\theta +\alpha )\sin \alpha \] \[=3\sin (\theta +\alpha )\cos \alpha +3\cos (\theta +\alpha )\sin \alpha \] \[\Rightarrow \,-2\sin (\theta +\alpha )\cos \alpha =4\cos (\theta +\alpha )sin\alpha \] \[\Rightarrow \frac{-\sin (\theta +\alpha )}{\cos (\theta +\alpha )}=\frac{2\sin \alpha }{\cos \alpha }\Rightarrow \tan (\theta +\alpha )+2\tan \alpha =0\]
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