A) \[1/4\]
B) \[1/2\]
C) \[1\]
D) \[4\]
Correct Answer: A
Solution :
\[\sin A.\sin (60{}^\circ -A)sin(60{}^\circ +A)=k\,sin3A\] \[\Rightarrow \,\,\,\sin A.\frac{\sin 3A}{4\sin A}=k.\sin 3A\] \[\left[ \because \,\,\,\sin (60{}^\circ +A).sin(60{}^\circ -A)=\frac{\sin 3A}{4\sin A} \right]\] \[\Rightarrow \,\,\,\frac{\sin 3A}{4}=k.\sin 3A\] \[\therefore \,k=\frac{1}{4}\]
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