Hi Kumar!

dy/dx= cos(x+y)                    =>(1)

let x+y = v

then, 1 +dy/dx = dv/dx

dy/dx = dv/dx-1

Put x+y = v and dy/dx = dv/dx - 1   in  eq. (1)

dv/dx - 1 = cosv

dv/dx = cos v +1

(1/(1+cosv)).dv=dx

(1/2)sec^2(v/2).dv=dx

Integrating both sides we get tan(v/2)=x+C

Thus tan((x+y)/2) = x +C is solution.