Minimum value of $\frac{{\sec }^4\mathrm{\alpha }}{{\tan }^2\mathrm{\beta }}+\frac{{\sec }^4\mathrm{\beta }}{{\tan }^2\mathrm{\alpha }},\text{ where }\mathrm{\alpha }\ne \frac{\mathrm{\pi }}{2}$,$\mathrm{\beta }\ne \frac{\mathrm{\pi }}{2},0<\mathrm{\alpha },\mathrm{\beta }<\frac{\mathrm{\pi }}{2}$ is

Let $\mathrm{a}={\tan }^2\mathrm{\alpha },\mathrm{b}={\tan }^2\mathrm{\beta }$

Given expression becomes

$\begin{array}{l}\frac{(\mathrm{a}+1{)}^2}{\mathrm{b}}+\frac{(\mathrm{b}+1{)}^2}{\mathrm{a}}\\ =\frac{{\mathrm{a}}^2+2\mathrm{a}+1}{\mathrm{b}}+\frac{{\mathrm{b}}^2+2\mathrm{b}+1}{\mathrm{a}}\\ =\frac{{\mathrm{a}}^2}{\mathrm{b}}+\frac{1}{\mathrm{b}}+\frac{{\mathrm{b}}^2}{\mathrm{a}}+\frac{1}{\mathrm{a}}+2\left(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}\right)\\ \ge 4\sqrt[4]{\frac{{\mathrm{a}}^2}{\mathrm{b}}\cdot \frac{1}{\mathrm{b}}\cdot \frac{{\mathrm{b}}^2}{\mathrm{a}\cdot \mathrm{a}}}+2\left(2\right)\\ =4+4=8\end{array}$