Question

In triangle ABC, prove the following:
$b\sin B-c\sin C=a\sin \left(B-C\right)$

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
Then,

Consider the LHS of he equation $b\sin B-c\sin C=a\sin \left(B-C\right)$.
LHS$=k\sin B\sin B-k\sin C\sin C$
$$\begin{gathered} =k\left({\sin }^{2}B-{\sin }^{2}C\right) \\ =k\left[\sin \left(B+C\right)\sin \left(B-C\right)\right]\left[\because {\sin }^2\mathrm{B}-{\sin }^2\mathrm{C}=\sin \left(\mathrm{B}+\mathrm{C}\right)\sin \left(\mathrm{B}-\mathrm{C}\right)\right] \\ =k\left[\sin \left(\mathrm{\pi }-\mathrm{A}\right)\sin \left(B-C\right)\right]\left[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right] \\ =k\sin A\sin \left(B-C\right)\left[\because a=k\sin A\right] \\ =a\sin \left(B-C\right)=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$