If $\mathrm{x}=\mathrm{Cot\theta } + \mathrm{tan\theta };\mathrm{y}=\mathrm{sec\theta }-\mathrm{cos\theta }$  $then {\left({\mathrm{x}}^2\mathrm{y}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\left({\mathrm{xy}}^2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=$

$$\begin{gathered} \mathrm{x}=\mathrm{cot\theta }+\mathrm{tan\theta } ; \mathrm{y}=\mathrm{sec\theta }-\mathrm{cos\theta } \\ \mathrm{x}=\frac{1}{\mathrm{sin\theta } \mathrm{cos\theta }} \mathrm{y}=\frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }} \\ {\mathrm{x}}^2\mathrm{y}=\frac{1}{{\sin }^2\mathrm{\theta } {\cos }^2\mathrm{\theta }}\frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }} \\ {\mathrm{x}}^2\mathrm{y}={\sec }^3\mathrm{\theta } \\ \mathrm{x} {\mathrm{y}}^2=\frac{1}{\mathrm{som\theta } \mathrm{cos\theta }}.\frac{{\sin }^4\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}={\mathrm{Tan}}^3\mathrm{\theta } \\ {\left({\mathrm{x}}^2\mathrm{y}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{(\mathrm{x} {\mathrm{y}}^2)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} ={\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \end{gathered}$$