If $0<\alpha ,\beta <\pi$ and $\cos \alpha +\cos \beta -\cos (\alpha +\beta )=3/2$ then $\sin \alpha +\cos \beta$ is  equal to

From the given equation we have

$\begin{array}{r}2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-2{\cos }^2\frac{\alpha +\beta }{2}+1=\frac{3}{2}\\ \Rightarrow 4{\cos }^2\frac{\alpha +\beta }{2}-4\cos \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}+1=0\end{array}$

$\Rightarrow {\left(2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2}\right)}^2={\cos }^2\frac{\alpha -\beta }{2}-1 \left(i\right)$

so the only possibility is ${\cos }^2\frac{\alpha -\beta }{2}-1=0$

since ${\cos }^2\left(\frac{\alpha -\beta }{2}\right)\le 1$

As $0<\alpha ,\beta <\pi$, we have $\alpha =\beta$

From (i) we get $\cos \alpha =1/2=\cos \beta$ and $\sin \beta =\sin \alpha =\frac{\sqrt{3}}{2}$. so that $\sin \alpha +\cos \beta =\frac{\sqrt{3}+1}{2}$