If A + B + C = π, prove that

Taking L.C.M

Multiplying and divide the above equation by 2, we get

Since , sin2A = 2sinAcosA

NOW,

= sin2A + sin2B + sin2C

= 2sinAcosA + 2sin(B+C)cos(B - C)
since A + B + C = π

= 2sinAcosA + 2sin(π - A)cos(B - C )
= 2sinAcosA + 2sinAcos(B - C)
= 2sinA{cosA + cos (B-C)}
( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )

And now using
= 2sinA{2sinBsinC}
= 4sinAsinBsinC

Putting the above value in the equation, we get

= 2

= R.H.S