Trigogebra

$m^3=\frac{1}{\sin \alpha}-\sin \alpha=\frac{1-\sin^2\alpha}{\sin \alpha}=\frac{\cos^2\alpha}{\sin \alpha} \tag{1}$ $n^3=\frac{1}{\cos \alpha}-\cos \alpha=\frac{1-\cos^2\alpha}{\cos \alpha}=\frac{\sin^2\alpha}{\cos \alpha}\tag{2}$ Multiplying $(1)$ and $(2)$ , $m^3n^3=\frac{\cos^2\alpha}{\sin \alpha} \cdot \frac{\sin^2\alpha}{\cos \alpha}=\sin \alpha \cos\alpha$ $\implies m^2=\frac{\sin\alpha \cos\alpha}{mn^3}\;,\;\;\;n^2=\frac{\sin \alpha \cos \alpha}{m^3n}$ Dividing $(1)$ and $(2)$ , $\frac{m^3}{n^3}=\frac{\cos^2\alpha}{\sin \alpha}\cdot \frac{\cos\alpha}{\sin^2 \alpha} =\frac{\cos^3\alpha}{\sin^3\alpha}$ $\implies \frac{m}{n}=\frac{\cos\alpha}{\sin\alpha}$ Therefore, $\begin{aligned} m^2n^2(m^2+n^2)&=m^2n^2\left(\frac{\sin\alpha \cos\alpha}{mn^3}+\frac{\sin \alpha \cos \alpha}{m^3n}\right) \\ \\ &=\frac{m}{n}\sin\alpha\cos\alpha+\frac{n}{m}\sin\alpha\cos\alpha \\ \\ &=\frac{\cos\alpha}{\sin\alpha}\cdot\sin\alpha\cos\alpha+\frac{\sin\alpha}{\cos\alpha}\cdot \sin\alpha\cos\alpha \\ \\ &=\cos^2\alpha+\sin^2\alpha \\ \\ &=\boxed{1} \end{aligned}$

$\begin{aligned} m^3 & = \frac 1{\sin \alpha} - \sin \alpha = \frac {1-\sin^2 \alpha}{\sin \alpha} = \frac {\cos^2 \alpha}{\sin \alpha} \\ n^3 & = \frac 1{\cos \alpha} - \cos \alpha = \frac {1-\cos^2 \alpha}{\cos \alpha} = \frac {\sin^2 \alpha}{\cos \alpha} \\ \implies m^3n^3 & = \frac {\cos^2 \alpha}{\sin \alpha} \cdot \frac {\sin^2 \alpha}{\cos \alpha} = \sin \alpha \cos \alpha \\ \frac {m^3}{n^3} & = \frac {\cos^2 \alpha}{\sin \alpha} \cdot \frac {\cos \alpha}{\sin^2 \alpha} = \frac {\cos^3 \alpha}{\sin ^3 \alpha} \\ \implies \frac mn & = \frac {\cos \alpha}{\sin \alpha} \\ m^2n^2(m^2+n^2) & = m^4n^2 + m^2n^4 \\ & = m^3n^3 \cdot \frac mn + m^3n^3 \cdot \frac nm \\ & = \sin \alpha \cos \alpha \cdot \frac {\cos \alpha}{\sin \alpha} + \sin \alpha \cos \alpha \cdot \frac {\sin \alpha}{\cos \alpha} \\ & = \cos^2 \alpha + \sin^2 \alpha \\ & = \boxed 1 \end{aligned}$