Correct Answer - `-log_(e)|1-tan(x+y)/(2)|=x+c`
`(dy)/(dx) = cos(x+y)-sin(x+y)`
Putting `x+y=t`, we get `(dy)/(dx)=(dt)/(dx)-1`
Therefore, `(dt)/(dx)-1=cost-sint`
or `(dt)/(1+cost-sint)=dx` or `(sec^(2)t/2dt)/(2(1-tant/2))=dx`
or `-"ln "|1-tan(x+y)/(2)|=x+c`
