Question

a+b+c=0 then prove that sin2a+sin2b+sin2c=-4sina.sinb.sinc​

Answer 1

Answer:

A+B+C=180LHS=sin2A+sin2B+sin2C=2sin(A+B)cos(A−B)+2sinCcosC=2sinCcos(A−B)+2sinCcosC=2sinC(cos(A−B)+cosC)=2sinC(cos(A−B)−cos(A+B))=2sinC2sinAsinB=4sinAsinBsinC=RHS mere id ko follow kare