To prove:
\(sin(\frac{B-C}{2})=\frac{b-c}{a}.cos\frac{A}{2}\)
Or, \(\frac{b-c}{a}=\frac{sin(\frac{B-C}{2})}{cos\frac{A}{2}}\)
L.H.S. = \(\frac{b-c}{a}=\frac{k\,sin\,B\,-\,k\,sin\,C}{k\,sin\,A}\)
\(=\frac{sin\,B\,-\,sin\,C}{sin\,A}\)
\(=\frac{2\,cos(\frac{B+C}{2})\,sin\,(\frac{B-C}{2})}{2\,sin\frac{A}{2}\,cos\frac{A}{2}}\)
Now, in a triangle,
\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90°\)
Hence,
\(\frac{A}{2}=90°-(\frac{B+C}{2})\)
∴ LHS = RHS