In `DeltaABC`, prove that:
`sin(A+B/2).cosB/2=(c+a)/(a+b)cosC/2.cos(A-B)/(2)`

Explanation

RHS. `=(c+a)/(a+b).cosC/2.cos(A-B)/(2)`
`(ksinc+ksinA)/(ksinA+ksinB).cosC/2cos(A-B)/(2)`
`=(2sin(C+A)/(2).cos(C-A)/(2))/(2ain(A+B)/(2).cos(A-B)/(2)).cosC/2.cos(A-B)/(2)`
`=(sin(180^(@)-B)/(2).cos(180^(2)-(A+B)-A)/(2))/(sin(180^(@)-C)/(2)).cos(C/2)`
`=(sin(90^(@)-B/2)cos{90^(@)-(A+B/2)})/(sin(90^(@)-C/2)cosC/2`
`=(cosB/2.sin(A+B/2))/(cos(C/2)).cos(C/2)`
`=cos(B/2).sin(A+B/2)`=LHS Hence Proved.