It seems to me that before I tacklyethis I need to work out what the restrictions are on alpha and beta.
I am just going to elaborate on your first method first :)
I wish you (the question asker and answerer) were a member because then I would be much more sure that you would at least see I have done another answer :(
Yes, I know you already have the answer. 
I still want you to see mine though 
\(Given : \\sin\alpha+sin\beta=1 , \\cos\alpha+cos\beta=1\\ \text{Find the value of }sin\alpha-cos\beta\\ \\~\\ \text{since ALL sin and cos values must be between -1 and +1 AND} \\sin\alpha+sin\beta=1 \;\;\\\text{It follows that }sin\alpha \;\;and\;\; sin\beta \text { must both be between 0 and 1 }\\ so\;\; \alpha \;and\; \beta \;\text{must both be in the first or second quadrant}\\ cos\alpha+cos\beta=1 \\\text{It follows that }cos\alpha \;\;and\;\; cos\beta \text { must both be between 0 and 1 }\\ so\;\; \alpha \;and\; \beta \;\text{must both in the first quadrant}\\ \text{HENCE: }sin\alpha, \;\;sin\beta\; \;cos\alpha,\; cos\beta\; \text {are all between 0 and 1} \)
Squaring both original equations, which we can do becasue everything is positive we have:
\(sin^2\alpha+sin^2\beta+2sin\alpha sin\beta=1\\ cos^2\alpha+cos^2\beta+2cos\alpha\cos\beta=1\\ add\\ 1+1+2(sin\alpha sin\beta+cos\alpha cos\beta)=2\\ 2(sin\alpha sin\beta+cos\alpha cos\beta)=0\\ sin\alpha sin\beta+cos\alpha cos\beta=0\\ cos(\alpha-\beta)=cos(\beta-\alpha)=0 \quad Let\;\beta\ge \alpha \quad (I\;mean\;the\; 0 \;\;to\;\; \frac{\pi}{2}\; \text {equivalent angle)}\\ cos(\beta-\alpha)=0\\ \color{blue}{\text{From here on I am going to use n as an integer but its value can be different each time}}\\ \beta-\alpha=\frac{\pi}{2}+2\pi n \qquad n\in Z\\ \beta=\alpha+\frac{\pi}{2}+2\pi n \qquad n\in Z\\ \text{But }\beta \text{ must be in the first quadrant so }\alpha = 0+2\pi n \;\;and\;\;\beta=\frac{\pi}{2}+2\pi n \\ so\\ sin\alpha - \cos\beta=sin(2\pi n)- cos(\frac{\pi}{2}+2\pi n)\\ sin\alpha - \cos\beta=0- 0\\ sin\alpha - \cos\beta=0\\ \)
I think that method will stand up to scrutiny. I haven't evewn looked at your second answer yet :)
+529 
