If 0 < x,y < π and cosx - cos(x+y) = 3/4,

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$

$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}\right)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow\left(\cos \left(\frac{x+y}{2}\right)-\frac{1}{2} \cos \left(\frac{x-y}{2}\right)\right)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and

$\cos \left(\frac{x+y}{2}\right)=\frac{1}{2} \cos \left(\frac{x-y}{2}\right)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$\therefore \quad \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

option (2)