Let α and β be real numbers such that - π 4 < β < 0 < α < π 4 …

${\left(\frac{\sin \alpha }{\cos \beta }+\frac{\cos \alpha }{\sin \beta }+\frac{\cos \beta }{\sin \alpha }+\frac{\sin \beta }{\cos \alpha }\right)}^2$

$={\left(\frac{\cos \left(\alpha -\beta \right)}{\sin \beta \cos \beta }+\frac{\cos \left(\alpha -\beta \right)}{\sin \alpha \cos \alpha }\right)}^2$

$={\left(2\cos \left(\alpha -\beta \right)\left\{\frac{1}{\sin 2\beta }+\frac{1}{\sin 2\alpha }\right\}\right)}^2$

$=\frac{16}{9}{\left(\frac{2\sin \left(\alpha +\beta \right)·\cos \left(\alpha -\beta \right)}{\sin 2\alpha ·\sin 2\beta }\right)}^2$

$=\frac{16}{9}{\left(\frac{4·\frac{1}{3}·\frac{2}{3}}{\cos \left(2\alpha -2\beta \right)-\cos \left(2\alpha +2\beta \right)}\right)}^2$

$=\frac{16}{9}{\left(\frac{\frac{8}{9}}{\left(2{\cos }^2\left(\alpha -\beta \right)-1\right)-\left(1-2{\sin }^2\left(\alpha +\beta \right)\right)}\right)}^2$

$=\frac{16}{9}{\left(\frac{\frac{8}{9}}{\frac{8}{9}-2+\frac{2}{9}}\right)}^2$

$=\frac{16}{9}$

i.e. $\left[\frac{16}{9}\right]=1$