If `A+B+C=pi,` prove that `sin^2A/2+sin^2B/2-sin^2C/2=1-2cosA/2cosB/2sinC/2dot`

Explanation

LHS`=sin^(2)A/2+sin^(2)B/2- sin^(2)C/2`
`=(1-cosA)/(2) + (1-cosB)/(2) - (1-cosC)/(2)`
`=1/2-1/2(cosA+cosB-cosC)`…………..(i)
Now, `cosA+ cosB-cosC = 2cos(A+B)/(2) cos(A-B)/(2)- cosC`
`=2cos(pi-C)/(2) cos(A-B)/2-(1-2sin^(2)C/2)`
`2sinC/2{cos(A-B)/(2) + cos(A+B)/(2)}-1`
`=4 cosA/2 . cosB/2. sinC/2-1`
Now, (i) gives
L.H.S `=1/2-1/2[4 cosA/2 cosB/2sinC/2-1]`
`=1-2cosA/2 cosB/2 sinC/2`