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prove that `a^2sin2B+b^2sin2A=4Delta`

Answer» `L.H.S. = a^2sin2B+b^2sin2A`
`= (2RsinA)^2(2sinBcosB)+(2RsinB)^2(2sinAcosA)`
`=8R^2sinAsinB(sinAcosB+sinBcosA)`
`=8R^2sinAsinB(sin(A+B))`
`=8R^2sinAsinB(sin(pi-C))`
`=8R^2sinAsinBsinC`
Now, we know, `Delta = 2R^2sinAsinBsinC `
`:. 8R^2sinAsinBsinC = 4Delta = R.H.S.`


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