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.
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