If $\cot \alpha +\tan \alpha =m\text{ and }\frac{1}{\cos \alpha }-\cos \alpha$ $=n,$ then 

Clearly $\alpha \ne 0$

$\Rightarrow \begin{array}{r}\cot \alpha +\tan \alpha =m\Rightarrow 1+{\tan }^2\alpha =m\tan \alpha \\ {\sec }^{\mathrm{\prime }}\alpha =m\tan \alpha ( 1 )\end{array}$

and  $\frac{1}{\cos \alpha }-\cos \alpha =n\Rightarrow {\sec }^2\alpha -1=n\sec \alpha$

 $\begin{array}{ll}\Rightarrow & {\tan }^2\alpha =n\sec \alpha \Rightarrow {\tan }^4\alpha =n^2{\sec }^2\alpha \\ \Rightarrow & {\tan }^4\alpha =n^{2}m\tan \alpha \end{array}[\text{ by (1)] }$

$\Rightarrow {\tan }^3\alpha =n^{2}m\Rightarrow \tan \alpha ={\left(n^{2}m\right)}^{1/3}$

and  ${\sec }^2\alpha =m{\left(n^{2}m\right)}^{1/3} by ( 1 )$

Now, ${\sec }^2\alpha -{\tan }^2\alpha =1\Rightarrow m{\left(n^{2}m\right)}^{1/3}-{\left(n^{2}m\right)}^{2/3}=1$

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