If $\cos ec\theta -\sin \theta =m$ and $\sec \theta -\cos \theta =n$ then   $m^{\frac{4}{3}}.n^{\frac{2}{3}}+m^{\frac{2}{3}}.n^{\frac{4}{3}}$is

$\cos ec\theta -\sin \theta =m \Rightarrow \frac{1}{\sin \theta }-\sin \theta =m$

$\Rightarrow \frac{1-{\sin }^2\theta }{\sin \theta }=m\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }=m$……. (1)

$\sec \theta -\cos \theta =n$$\Rightarrow \frac{{\sin }^2\theta }{\cos \theta }=n$

$\left(1\right) \Rightarrow \sin \theta =\frac{{\cos }^2\theta }{m}, \left(2\right)\Rightarrow \frac{{\cos }^4\theta }{m^2\cos \theta }=n\Rightarrow {\cos }^3\theta =m^{2}n$

$\cos \theta ={\left(m^{2}n\right)}^{1/3}\Rightarrow {\cos }^2\theta ={\left(m^{2}n\right)}^{2/3},{\sin }^2\theta ={\left(m{n}^2\right)}^{2/3}$

$\therefore {\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}=1$