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Question

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

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Solution

## 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

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