Question

For any triangle ABC prove sin(B-C)/sin(B+C)=b²-c²/a².

Answer 1

i) Multiplying numerator and denominator by sin(B + C)
sin(B - c)/sin(B + C) = {sin(B - C)*sin(B + C)}/sin²(B + C) --------- (1)

ii) In a triangle, by angle sum property, A + B + C = 180 deg
==> B + C = 180 - A; so sin(B + C) = sin(180 - A) = sin(A)
Hence sin²(B + C) = sin²A ------- (2)

sin(B + C)*sin(B - C) = (sinBcosC + cosBsinC)(sinBcosC - cosBsinC) = sin²B*cos²C - cos²B*sin²C
Transferring cos in the form of sin and simplifying,
sin(B + c)*sin(B - C) = sin²B - sin²C -------- (3)

iii) So substituting the values of (2) & (3)in (1),
sin(B - C)/sin(B + C) = (sin²B - sin²C)/sin²A

By sine law of triangles, sin(A) = k*a; sinB = k*b and sinC = k*c
So, sin(B - C)/sin(B + C) = k²(b² - c²)/(k²*a²) = (b² - c²)/a²

Thus it is proved that, sin(B - c)/sin(B + C) = (b² - c²)/a²