b2sin2C +c2sin2B= 2bcsinA
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b²sin2C+c²sin2B
=b²(2sinCcosC)+c²(2sinBcosB)
=2bcosC.bsinC+2ccosB.csinB
=2bcosC.csinB+2ccosB.csinB
[∵, b/sinB=c/sinC; ∴, bsinC=csinB]
=2csinB(bcosC+ccosB)
=2csinB.a [∵, bcosC+ccosB=a]
=2ac. (bsinA/a)
[∵, a/sinA=b/sinB; ∴, asinB=bsinA; or, sinB=(bsinA/a)]
=2bcsinA (Proved)
=b²(2sinCcosC)+c²(2sinBcosB)
=2bcosC.bsinC+2ccosB.csinB
=2bcosC.csinB+2ccosB.csinB
[∵, b/sinB=c/sinC; ∴, bsinC=csinB]
=2csinB(bcosC+ccosB)
=2csinB.a [∵, bcosC+ccosB=a]
=2ac. (bsinA/a)
[∵, a/sinA=b/sinB; ∴, asinB=bsinA; or, sinB=(bsinA/a)]
=2bcsinA (Proved)
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