Question

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

Answer 1

$ \alpha \in\left(0, \frac{\pi}{4}\right), \beta \in\left(-\frac{\pi}{4}, 0\right) \Rightarrow \alpha+\beta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) $
$ \sin (\alpha+\beta)=\frac{1}{3}, \cos (\alpha-\beta)=\frac{2}{3} $
$ \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 (\alpha-\beta)}{\cos \beta \sin \beta}+\frac{\cos (\beta-\alpha)}{\sin \alpha \cos \alpha}\right)^2 $
$ =4 \cos ^2(\alpha-\beta)\left(\frac{1}{\sin 2 \beta}+\frac{1}{\sin 2 \alpha}\right)^2 $
$ =4 \cos ^2(\alpha-\beta)\left(\frac{2 \sin (\alpha+\beta) \cos (\alpha-\beta)}{\sin 2 \alpha \sin 2 \beta}\right).....$(1)
$ =\frac{16 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta) \times 4}{(\cos 2(\alpha-\beta)-\cos 2(\alpha+\beta))^2} $
$ =\frac{64 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta)}{\left(2 \cos ^2(\alpha-\beta)-1-1+2 \sin ^2(\alpha+\beta)\right)^2} $
$ =64 \times \frac{16}{81} \times \frac{1}{9} \frac{1}{\left(2 \times \frac{4}{9}-1-1+\frac{2}{9}\right)^2}$
$ =\frac{64 \times 16}{81 \times 9} \cdot \frac{81}{64}=\frac{16}{9} $
$ {\left[\frac{16}{9}\right]=1 \text { Ans. }}$