If `sinA+sinB=a `and `cosA+cosB=b` then prove that `sin(A+B)=(2ab)/(a^2+b^2)` and `cos(A+B)=(b^2-a^2)/(a^2+b^2)`

Explanation

`b^(2)+a^(2)=(cosA+cosB)^(2)+(sinA+sinB)^(2)`
`=cos^(2)A+cos^(2)B+2cosAcosB+sin^(2)A + sin^(2)B+2sinAsinB`
`=2+2cos(A+B)`……………(1)
and `b^(2)-a^(2)=(cosA+cosB)^(2)-(sinA+sinB)^(2)`
`=(cos^(2)A+cos^(2)B+2cosAcosB)-sin^(2)A+sin^(2)B+2sinAsinB`
`=cos(A-sin^(2)B)+(cos^(2)B-sin^(2)A)+2cos(B+A)cos(B-A)+2cos(A+B)`
`=cos(A+B)[cos(A-B)+cos(A-B)+2]`
`=cos(A+B)(b^(2)+a^(2))` [From eq.(1)]
`rArr cos(A+B)=(b^(2)-a^(2))/(b^(2)+a^(2))` Hence Proved.