In `triangle ABC` , with the usual notations , prove that `2{a sin^(2).(C)/(2)+csin^2.(A)/(2)}=a+c-b`

Explanation

L.H.S.= `2{asin^(2).(C)/(2)+csin^(2).(A)/(2)}`
`=a{2sin^(2).(C)/(2)}+c{2sin^(2).(A)/(2)}`
`=a(1-cos C)+c(1-cosA)`
`=a[1-(a^(2)+b^(2)-c^(2))/(2ab)]+c[1-(b^(2)+c^(2)-a^(2))/(2bc)]`
`=a[(2ab-a^(@)-b^(2)+c^(2))/(2ab)]+c[(2bc-b^(2)-c^(2)+a^(2))/(2bc)]`
`=(2ab-a^(2)-b^(2)+c^(2)+2ab-b^(2)-c^(2)+a^(2))/(2b)`
`=(2ab+2bc-2b^(2))/(2b)`
`=(2b(a+c-b))/(2b)=a+c-b=R.H.S.`