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

$x+y=\frac{{\sin }^{5}p+{\cos }^{5}p}{{\sin }^{2}p{\cos }^{2}p}$

$\begin{array}{l}=\frac{{\sin }^{3}p\left(1-{\cos }^{2}p\right)+{\cos }^{3}p\left(1-{\sin }^{2}p\right)}{{\sin }^{2}p{\cos }^{2}p}\\ =\frac{{\sin }^{3}p+{\cos }^{3}p-{\sin }^{2}p{\cos }^{2}p(\sin p+\cos p)}{{\sin }^{2}p{\cos }^{2}p}\end{array}$

$=\frac{(\sin p+\cos p{)}^3-3\sin p\cos p(\sin p+\cos p)-}{{\sin }^{2}p{\cos }^{2}p(\sin p+\cos p)}$

Now $2\sin p\cos p=(\sin p+\cos p{)}^2-1=-3/4$

$x+y=\frac{{\left(\frac{1}{2}\right)}^3-3\left(-\frac{3}{8}\right)\left(\frac{1}{2}\right)-{\left(-\frac{3}{8}\right)}^2\left(\frac{1}{2}\right)}{{\left(-\frac{3}{8}\right)}^2}=\frac{79}{18}$