Let $\alpha$ and $\beta$ be real number 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 {(\frac{cos\alpha }{sin\beta }+\frac{sin\alpha }{cos\beta }+\frac{sin\beta }{cos\alpha }+\frac{cos\beta }{sin\alpha })}^2is _________________ .

The correct answer is 1.

We are given the trigonometric values:

$sin(\alpha +\beta )=\frac{1}{3}$

$cos(\alpha -\beta )=\frac{2}{3}$

Let us denote the given expression inside the bracket as S:

$S=\frac{cos\alpha }{sin\beta }+\frac{sin\alpha }{cos\beta }+\frac{sin\beta }{cos\alpha }+\frac{cos\beta }{sin\alpha }$

We can group the terms to simplify S:

$S=\left(\frac{cos\alpha }{sin\beta }+\frac{cos\beta }{sin\alpha }\right)+\left(\frac{sin\alpha }{cos\beta }+\frac{sin\beta }{cos\alpha }\right)$

Combining the fractions in each bracket:

$S=\frac{sin\alpha cos\alpha +sin\beta cos\beta }{sin\alpha sin\beta }+\frac{sin\alpha cos\alpha +sin\beta cos\beta }{cos\alpha cos\beta }$

Using the double-angle formula $sin2x=2sinxcosx$, the numerator becomes:

$sin\alpha cos\alpha +sin\beta cos\beta =\frac{1}{2}\left(sin2\alpha +sin2\beta \right)$

Factoring out the common numerator gives:

$S=\frac{1}{2}\left(sin2\alpha +sin2\beta \right)\left(\frac{1}{sin\alpha sin\beta }+\frac{1}{cos\alpha cos\beta }\right)$

$S=\frac{1}{2}\left(sin2\alpha +sin2\beta \right)\left(\frac{cos\alpha cos\beta +sin\alpha sin\beta }{sin\alpha cos\alpha sin\beta cos\beta }\right)$

Using the identity $cos(\alpha -\beta )=cos\alpha cos\beta +sin\alpha sin\beta$ and expressing the denominator in terms of double angles:

Step 1: Calculate $sin2\alpha +sin2\beta$

Using the sum-to-product formula:

$sin2\alpha +sin2\beta =2sin(\alpha +\beta )cos(\alpha -\beta )=2\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)=\frac{4}{9}$

Step 2: Calculate $sin2\alpha sin2\beta$

We know:

$cos(2\alpha -2\beta )=2{cos}^2(\alpha -\beta )-1=2{\left(\frac{2}{3}\right)}^2-1=\frac{8}{9}-1=-\frac{1}{9}$

$cos(2\alpha +2\beta )=1-2{sin}^2(\alpha +\beta )=1-2{\left(\frac{1}{3}\right)}^2=1-\frac{2}{9}=\frac{7}{9}$

Using the identity $cos(A-B)-cos(A+B)=2sinAsinB$:

$2sin2\alpha sin2\beta =-\frac{1}{9}-\frac{7}{9}=-\frac{8}{9}\Rightarrow sin2\alpha sin2\beta =-\frac{4}{9}$

Step 3: Substitute back into S

$S=\frac{1}{2}\left(\frac{4}{9}\right)·\frac{\frac{2}{3}}{\frac{1}{4}\left(-\frac{4}{9}\right)}$

$S=\frac{2}{9}·\frac{\frac{2}{3}}{-\frac{1}{9}}=\frac{2}{9}·\left(-6\right)=-\frac{4}{3}$

Step 4: Compute $S^2$ and find the greatest integer

$S^2={\left(-\frac{4}{3}\right)}^2=\frac{16}{9}\approx 1.777$

The greatest integer less than or equal to $\frac{16}{9}$ is:

$\left[\frac{16}{9}\right]=1$