If A+B+C= π than prove that sin2A+sin2B-sin2C=4.cosA.cosB.sinC

Steps to prove sin2A+sin2B-sin2C=4.cosA.cosB.sinC

L.H.S

=sin2A+sin2B-sin2C

=2sin(2A+2B)/2.cos(2A-2B)/2-2.sinC.cosC

=2sin(A+B).cos(A-B)-2.sinC.cosC

=2sinC{cos(A-B)-cosC} [sin(A+B)=sinC]

=2sinC{cos(A-B)+cos(A+B)} [cos(A+B)=-cosC]

=2.sinC.2.cosA.cosB

=4.cosA.cosB.cosC

=R.H.S

Hence sin2A+sin2B-sin2C=4.cosA.cosB.sinC is proved.

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