If $x=\frac{{\sin }^{3}p}{{\cos }^{2}p},y=\frac{{\cos }^{3}p}{{\sin }^{2}p};and\sin p+\cos p=\frac{1}{2},then x+y=$

 

$\begin{array}{l}x+y=\frac{{\sin }^{5}p+{\cos }^{5}p}{{\sin }^{2}p{\cos }^{2}p}=\frac{{\sin }^{3}p(1-{\cos }^{2}p)+{\cos }^{3}p(1-{\sin }^{2}p)}{{\sin }^{2}p{\cos }^{2}p}\\ =\frac{{(\sin p+\cos p)}^3-3\sin p\cos p(\sin p+\cos p)-{\sin }^{2}p{\cos }^{2}p(\sin p+\cos p)}{{\sin }^{2}p{\cos }^{2}p}\\ Now 2\sin p\cos p={(\sin p+\cos p)}^2-1=-3/4\\ \therefore x+y=\frac{{(1/2)}^3-3(-3/8)(1/2)-{(-3/8)}^2(1/2)}{{(-3/8)}^2}=\frac{79}{18}.\end{array}$