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Question
In ΔABC with usual notations, prove that 2a `{sin^2(C/2)+csin^2 (A/2)}` = (a + c - b)
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Solution
`sin^2theta =(1-cos2theta)/2`
`L.H.S=2{asin^2(C/2)+csin^2(A/2)}`
`=2{(a(1-cosC))/2+(c(1-cosA))/2}`
=a-acosC+c-ccosA
=(a+c)-(acosC+ccosA)
=a+c-b
R.H.S
2a `{sin^2(C/2)+csin^2 (A/2)}` = (a + c - b)
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