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

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