A) 0
B) \[a+b+c\]
C) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
D) \[2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]
Correct Answer: A
Solution :
\[a\sin (B-C)+b\sin (C-A)+c\sin (A-B)\] \[=k\,(\Sigma \sin A\sin (B-C)\]\[=k\,\{\Sigma \sin (B+C)\sin (B-C)\}\] \[=k\,\left\{ \Sigma \frac{1}{2}(\cos 2C-\cos 2B) \right\}=0\]. Note: Students should note here that most of the expressions containing the cyclic factor associating with ?-? reduces to 0.
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