Let 𝛼 and 𝛽 be real numbers such that $-\frac{\mathrm{\pi }}{4}<\mathrm{\beta }<0<\mathrm{\alpha }<\frac{\mathrm{\pi }}{4}$. If $\sin (\mathrm{\alpha }+\mathrm{\beta })=\frac{1}{3}$ and $\cos (\mathrm{\alpha }-\mathrm{\beta })=\frac{2}{3}$ then the greatest integer less than or equal to $\frac{9}{4}{\left(\frac{\sin \mathrm{\alpha }}{\cos \mathrm{\beta }}+\frac{\cos \mathrm{\beta }}{\sin \mathrm{\alpha }}+\frac{\cos \mathrm{\alpha }}{\sin \mathrm{\beta }}+\frac{\sin \mathrm{\beta }}{\cos \mathrm{\alpha }}\right)}^2$is ________.

$\begin{array}{l}\frac{9}{4}{\left(\frac{\sin \mathrm{\alpha }}{\cos \mathrm{\beta }}+\frac{\cos \mathrm{\alpha }}{\sin \mathrm{\beta }}+\frac{\cos \mathrm{\beta }}{\sin \mathrm{\alpha }}+\frac{\sin \mathrm{\beta }}{\cos \mathrm{\alpha }}\right)}^2\\ =\frac{9}{4}{\left(\frac{\cos (\mathrm{\alpha }-\mathrm{\beta })}{\sin \mathrm{\beta cos}\mathrm{\beta }}+\frac{\cos (\mathrm{\alpha }-\mathrm{\beta })}{\sin \mathrm{\alpha }\cdot \cos \mathrm{\alpha }}\right)}^2\\ =\frac{9}{4}{\left(\frac{2}{3}\left\{\frac{1}{\sin 2\mathrm{\beta }}+\frac{1}{\sin 2\mathrm{\alpha }}\right\}\right)}^2\\ =\frac{9}{4}·\frac{4}{9}{\left(\frac{2\sin (\mathrm{\alpha }+\mathrm{\beta })\cdot \cos (\mathrm{\alpha }-\mathrm{\beta })}{\sin 2\mathrm{\alpha }\cdot \sin 2\mathrm{\beta }}\right)}^2\\ ={\left(\frac{1}{\cos (2\mathrm{\alpha }-2\mathrm{\beta })-\cos (2\mathrm{\alpha }+2\mathrm{\beta })}\right)}^2\\ ={\left(\frac{2}{2{\cos }^2(\mathrm{\alpha }-\mathrm{\beta })-1-1+2{\sin }^2(\mathrm{\alpha }+\mathrm{\beta })}\right)}^2\end{array}$
$\begin{array}{l}={\left(\frac{\frac{8}{9}}{\frac{8}{9}-2+\frac{2}{9}}\right)}^2\\ ={\left(1\right)}^2\\ =1\end{array}$