∵$$\frac{{{\sin }^{4}}\alpha }{{{\cos }^{2}}\beta }+\frac{{{\cos }^{4}}\alpha }{{{\sin }^{2}}\beta }=1$$，  即$$\frac{{{\sin }^{4}}\alpha }{{{\cos }^{2}}\beta }+\frac{{{\cos }^{4}}\alpha }{{{\sin }^{2}}\beta }-1={{\left( \frac{{{\sin }^{2}}\alpha }{\cos \beta }-\cos \beta \right)}^{2}}+{{\left( \frac{{{\cos }^{2}}\alpha }{\sin \beta }-\sin \beta \right)}^{2}}=0$$，  ∴$$\frac{{{\sin }^{2}}\alpha }{\cos \beta }-\cos \beta =0$$，$$\frac{{{\cos }^{2}}\alpha }{\sin \beta }-\sin \beta =0$$，  ∴$${{\sin }^{2}}\alpha ={{\cos }^{2}}\beta $$，$${{\cos }^{2}}\alpha ={{\sin }^{2}}\beta $$，  根据$$\alpha $$、$$\beta $$为锐角，可得$$\alpha +\beta =90{}^{}\circ $$．  故答案为：$$90{}^{}\circ $$．  由柯西不等式，$$\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)\left( \frac{{{\sin }^{4}}\alpha }{{{\cos }^{2}}\beta }+\frac{{{\cos }^{4}}\alpha }{{{\sin }^{2}}\beta } \right)\geqslant {{\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)}^{2}}=1$$，  因此，$$\frac{{{\cos }^{4}}\beta }{{{\sin }^{4}}\alpha }=\frac{{{\sin }^{4}}\beta }{{{\cos }^{4}}\alpha }$$，即$$\cos \alpha \cos \beta =\sin \alpha \sin \beta $$，$$\cos \left( \alpha +\beta \right)=0$$，故$$\alpha +\beta =\frac{ \pi }{2}$$．  故答案为：$$\frac{ \pi }{2}$$． 
已知锐角$$\alpha $$，$$\beta $$满足条件：$$\frac{{{\sin }^{4}}\alpha }{{{\cos }^{2}}\beta }+\frac{{{\cos }^{4}}\alpha }{{{\sin }^{2}}\beta }=1$$，则$$\alpha +\beta =$$~\uline{~~~~~~~~~~}~．
Options:
1. $$\frac{ \pi }{4}$$ 
2. $$\frac{ \pi }{2}$$ 
3. $$\frac{ 2\pi }{3}$$ 
4. $${ \pi }$$ 