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

Correct Answer: 1

\(\sin(\alpha + \beta) = \frac{1}{3}\) and  \(\cos(\alpha - \beta) = \frac{2}{3}\)

The expression to simplify is:

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

This expression is equal to:

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

This can be simplified further to:

\(= \left(\frac{4}{3} \left\{\frac{1}{\sin 2\beta} + \frac{1}{\sin 2\alpha}\right\}\right)^2\)

Next, simplifying more:

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

This becomes:

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

We simplify the denominator further:

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

This simplifies to:

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

Finally, we get:

\(= \frac{16}{9} = 1\)

Therefore, the correct answer is 1.