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Given alpha+beta+gamma=pi, prove that si...

Given `alpha+beta+gamma=pi,` prove that `sin^2alpha+sin^2beta-sin^2gamma=2sinalphasinbetacosgammadot`

Text Solution

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`LHS = sin^(2)alpha+sin^(2)beta-sin^(2)gamma`
`sin^(2)alpha+(sin^(2)beta-sin^(2)gamma)`
`=sin^(2)alpha + sin(beta+gamma)sin(beta-gamma)`
`=sin^(2)alpha + sin (pi -alpha)sin(beta - gamma)[because alpha+beta+gamma = pi]`
`=sin^(2)alpha+sinalphasin(beta-gamma)`
`=sin alpha [sinalpha + sin(beta-gamma)]`
`=sin alpha[sin (pi-(beta+gamma))+sin(beta-gamma)]`
`=sinalpha[sin(beta+gamma)+sin(beta-gamma)]`
`=sin alpha[2 sin beta cos gamma]`
`=2 sin alpha sin beta cos gamma = RHS`
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