in triangle ABC , prove that b^2 sin2C + c^2 sin2B = 2bc sinA

LHS of the given equation is:
b2sin2C+c2sin2B=b2.2sinCcosC+c2.2sinBcosB  [since sin2θ=2sinθ.cosθ=2.b2sinCcosC+c2sinBcosB=2.b2.Kc.a2+b2-c22ab+c2.Kb.a2+c2-b22ac   [by sine rule :sinCc=sinBb=sinAa=K=2Kbc.a2+b2-c22a+a2+c2-b22a     [by cosine rule :cosC=a2+b2-c22ab=2Kbc.a2+b2-c2+a2+c2-b22a=2Kbc.2a22a=2Kabc
=2bc.(Ka)=2bc.sinA=RHS

hope this helps you

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