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

Given :
\(\sin(\alpha+\beta)=\frac{1}{3}\) and \(\cos(\alpha-\beta)=\frac{2}{3}\)
\(\left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}\right)^2\)
\(=(\frac{\cos(\alpha-\beta)}{\sin\beta\cos\beta}+\frac{\cos(\alpha-\beta)}{\sin\alpha.\cos\alpha})^2\)
\(=\left(\frac{4}{3} \left\{\frac{1}{\sin2\beta}+\frac{1}{\sin2\alpha}\right\}\right)^2\)
\(=\frac{16}{9}\left(\frac{2\sin(\alpha+\beta).\cos(\alpha-\beta)}{\sin2\alpha.\sin2\beta}\right)^2\)
\(=\frac{16}{9}(\frac{4.\frac{1}{2}.\frac{2}{3}}{\cos(2\alpha-2\beta)-\cos(2\alpha+2\beta)})^2\)
\(=\frac{16}{9}\left(\frac{\frac{8}{9}}{2\cos^2(\alpha-\beta)-1-1+2\sin^2(\alpha+\beta)}\right)^2\)
\(=\frac{16}{9}(\frac{\frac{8}{9}}{\frac{8}{9}+2+\frac{2}{9}})\)
\(=\frac{16}{9}\)
\(=1\)
Therefore, the correct answer is 1.