If   ${\sin }^{4}x+{\cos }^{4}y+2=4$$\sin x\cos y,0\le x,y\le \pi /2$ then  $\sin x+\cos y=$

The given equation can be written as 
${\sin }^{4}x+{\cos }^{4}y+2-4\sin x\cos y=0$
$\Rightarrow {\left({\sin }^{2}x-1\right)}^2+{\left({\cos }^{2}y-1\right)}^2+2{\sin }^{2}x+2{\cos }^{2}y-4\sin x\cos y=0$
$\Rightarrow {\left({\sin }^{2}x-1\right)}^2+{\left({\cos }^{2}y-1\right)}^2+2{\left(\sin x-\cos y\right)}^2=0$
which is true if ${\sin }^{2}x=1,{\cos }^{2}y=1$ and  $\sin x=\cos y,$
so $\sin x+\cos y=2$as $0\le x,y\le \pi /2$.