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Prove that : 

\(sin(\frac{B-C}{2})=\frac{b-c}{a}.cos\frac{A}{2}\)

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Best answer

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

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