A) \[0\]
B) \[-2\]
C) \[1\]
D) \[1\]
Correct Answer: A
Solution :
Given that, \[\sin B=3\sin (2A+B)\] \[\Rightarrow \] \[\frac{\sin \,\,B}{\sin \,(2A+B)}=\frac{3}{1}\] \[\Rightarrow \] \[\frac{\sin B+\sin (2A+B)}{\sin B-\sin (2A+B)}=\frac{3+1}{3-1}\] (use componendo-dividendo formula) \[\Rightarrow \] \[-\frac{2\sin \,(A+B).cos(A)}{2\cos \,(A+B).\sin (A)}=\frac{4}{2}\] \[\Rightarrow \] \[\tan (A+B).\cot \,A=-2\] \[\Rightarrow \] \[\tan \,(A+B)=-2\tan A\] \[\Rightarrow \] \[2\tan A+\tan (A+B)=0\]
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