b2-c2dividedby a2=sin(B-C) divided by sin(B+C) - Maths - Solution of Triangles

Question

​b2-c​2dividedby
a2=sin(B-C) divided by sin(B+C)

Varun.Rawat · Accepted Answer

Let asin A = bsin B = csin C = k⇒a = k sin A; b = k sin B; c = k sin CNow, LHS = b2-c2a2=k2sin2B - k2sin2Ck2sin2A=sin2B - sin2Csin2A=sinB+C 
 sinB-Csin2A=sinπ-A 
 sinB-Csin2A    as, A + B + C = π=sin A 
 sinB-Csin2A=sinB-Csin A=sinB-Csinπ - B+C=sinB-CsinB+C    as, sinπ-θ = sin θ=RHS

​b2-c​2dividedby a2=sin(B-C) divided by sin(B+C)

b²-c²dividedby a²=sin(B-C) divided by sin(B+C)