Prove that : (b^2-c^2)/a^2 = [sin(B-C )]/[sin (B+C)]

Question

Prove that : \(\frac{b^2-c^2}{a^2}=\frac{sin(B-C)}{sin(B+C)}\)

Answer 1

LHS = \(\frac{b^2-c^2}{a^2}\)

= \(\frac{k^2\,sin^2\,B -k^2sin^2\,C}{k^2\,sin^2A}\)

= \(\frac{sin^2\,B -sin^2\,C}{sin^2A}\)

= \(\frac{sin\,(B+C) -sin(B-C)}{sin^2[180°-(B+C)]}\)

= \(\frac{sin\,(B+C) -sin(B-C)}{sin^2(B+C)}\)

= \(\frac{sin(B-C)}{sin(B+C)}\) = RHS

∴ LHS = RHS