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(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{

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$1$

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(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$.Grouping terms: $E = \left(\frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \beta \sin \beta} + \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \alpha}\right)^2$.Using $\cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,we get:$E = \cos^2(\alpha-\beta) \left(\frac{1}{\cos \beta \sin \beta} + \frac{1}{\sin \alpha \cos \alpha}\right)^2$.$E = \cos^2(\alpha-\beta) \left(\frac{2}{\sin 2\beta} + \frac{2}{\sin 2\alpha}\right)^2 = 4 \cos^2(\alpha-\beta) \left(\frac{\sin 2\alpha + \sin 2\beta}{\sin 2\alpha \sin 2\beta}\right)^2$.Using $\sin 2\alpha + \sin 2\beta = 2 \sin(\alpha+\beta) \cos(\alpha-\beta)$ and $\sin 2\alpha \sin 2\beta = \frac{1}{2}(\cos(2\alpha-2\beta) - \cos(2\alpha+2\beta)) = \cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)$:$E = 4 \cos^2(\alpha-\beta) \left(\frac{2 \sin(\alpha+\beta) \cos(\alpha-\beta)}{\cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)}\right)^2$.Substituting values: $\sin(\alpha+\beta) = \frac{1}{3}$,$\cos(\alpha-\beta) = \frac{2}{3}$.$E = 4 \left(\frac{4}{9}\right) \left(\frac{2 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\frac{4}{9} - \frac{1}{9}}\right)^2 = \frac{16}{9} \left(\frac{4/9}{3/9}\right)^2 = \frac{16}{9} \left(\frac{4}{3}\right)^2 = \frac{16}{9} \cdot \frac{16}{9} = \frac{256}{81} \approx 3.16$.Wait,re-evaluating the expression: $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha}\right) + \left(\frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right) = \frac{\sin^2 \alpha + \cos^2 \beta}{\sin \alpha \cos \beta} + \frac{\cos^2 \alpha + \sin^2 \beta}{\cos \alpha \sin \beta}$.Actually,the expression simplifies to $\frac{\cos(\alpha-\beta)}{\sin \alpha \cos \beta} + \frac{\cos(\alpha-\beta)}{\cos \alpha \sin \beta} = \cos(\alpha-\beta) \frac{\sin(\alpha+\beta)}{\sin \alpha \cos \alpha \sin \beta \cos \beta} = \frac{4 \cos(\alpha-\beta) \sin(\alpha+\beta)}{\sin 2\alpha \sin 2\beta}$.Using $\sin 2\alpha \sin 2\beta = \cos(2\alpha-2\beta) - \cos(2\alpha+2\beta) = 2\cos^2(\alpha-\beta) - 1 - (1 - 2\sin^2(\alpha+\beta)) = 2(\frac{4}{9}) - 2 + 2(\frac{1}{9}) = \frac{8}{9} - 2 + \frac{2}{9} = -\frac{8}{9}$.$E = \left(\frac{4 \cdot (2/3) \cdot (1/3)}{-8/9}\right)^2 = \left(\frac{8/9}{-8/9}\right)^2 = (-1)^2 = 1$.The greatest integer is $1$.

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