Answer :


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


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If A + B + C = π,RS Aggarwal - Mathematics

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