Question

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any triangle ABC, find the value of $a\sin \left(B-C\right)+b\sin \left(C-A\right)+c\sin \left(A-B\right)$.

Answer 1

Using sine rule, we have

$$\begin{gathered} a\sin \left(B-C\right)+b\sin \left(C-A\right)+c\sin \left(A-B\right) \\ =k\sin A\sin \left(B-C\right)+k\sin B\sin \left(C-A\right)+k\sin C\sin \left(A-B\right) \\ =k\sin \left[\mathrm{\pi }-\left(B+C\right)\right]\sin \left(B-C\right)+k\sin \left[\mathrm{\pi }-\left(C+A\right)\right]\sin \left(C-A\right)+k\sin \left[\mathrm{\pi }-\left(A+B\right)\right]\sin \left(A-B\right) \end{gathered}$$
$$\begin{gathered} =k\left[\sin \left(B+C\right)\sin \left(B-C\right)+\sin \left(C+A\right)\sin \left(C-A\right)+\sin \left(A+B\right)\sin \left(A-B\right)\right] \\ =k\left({\sin }^{2}B-{\sin }^{2}C+{\sin }^{2}C-{\sin }^{2}A+{\sin }^{2}A-{\sin }^{2}B\right) \\ =k\times 0 \\ =0 \end{gathered}$$

Hence, the required value is 0.