If $\cot \mathrm{\theta }+\tan \mathrm{\theta }=\mathrm{x}\text{ and }\sec \mathrm{\theta }-\cos \mathrm{\theta }=\mathrm{y}\text{, }$then 

$\begin{array}{l}\cot \mathrm{\theta }+\tan \mathrm{\theta }=\mathrm{x}\\ \mathrm{or} \frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}=\mathrm{x}\\ \mathrm{or} 1=\mathrm{xsin}\mathrm{\theta cos}\mathrm{\theta }\\ \mathrm{now}, \sec \mathrm{\theta }-\cos \mathrm{\theta }=\mathrm{y}\\ \frac{1}{\cos \mathrm{\theta }}-\cos \mathrm{\theta }=\mathrm{y}\\ \mathrm{or} {\sin }^2\mathrm{\theta }=\mathrm{ycos}\mathrm{\theta }\\ \mathrm{now} , {\mathrm{x}}^2\mathrm{y}=\frac{1}{{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }}\cdot \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }}=\frac{1}{{\cos }^3\mathrm{\theta }}\\ \mathrm{and} {\mathrm{xy}}^2=\frac{{\sin }^3\mathrm{\theta }}{{\cos }^3\mathrm{\theta }}\\ \therefore {\left({\mathrm{x}}^2\mathrm{y}\right)}^{2/3}-{\left({\mathrm{xy}}^2\right)}^{2/3}=\frac{1}{{\cos }^2\mathrm{\theta }}-\frac{{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}=1\end{array}$