Given, $$\sin (\alpha + \beta ) = {1 \over 3}$$

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

Let, $$E = {{\sin \alpha } \over {\cos \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\cos \alpha } \over {\sin \beta }} + {{\sin \beta } \over {\cos \alpha }}$$

$$ = {{\sin \alpha } \over {\cos \beta }} + {{\cos \alpha } \over {\sin \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\sin \beta } \over {\cos \alpha }}$$

$$ = {{\sin \alpha \sin \beta + \cos \alpha \cos \beta } \over {\sin \beta \cos \beta }} + {{\cos \alpha \cos \beta + \sin \alpha \sin \beta } \over {\sin \alpha \cos \alpha }}$$

$$ = {{\cos (\alpha - \beta )} \over {\sin \beta \cos \beta }} + {{\cos (\alpha - \beta )} \over {\sin \alpha \cos \alpha }}$$

$$ = \cos (\alpha - \beta )\left[ {{2 \over {2\sin \beta \cos \beta }} + {2 \over {2\sin \alpha \cos \alpha }}} \right]$$

$$ = {2 \over 3}\left[ {{2 \over {\sin 2\beta }} + {2 \over {\sin 2\alpha }}} \right]$$

$$ = {4 \over 3}\left[ {{1 \over {\sin 2\beta }} + {1 \over {\sin 2\alpha }}} \right]$$

$$ = {4 \over 3}\left[ {{{\sin 2\alpha + \sin 2\beta } \over {\sin 2\alpha \sin 2\beta }}} \right]$$

$$ = {{4 \times 2} \over 3}\left[ {{{2\sin \left( {{{2\alpha + 2\beta } \over 2}} \right)\cos \left( {{{2\alpha - 2\beta } \over 2}} \right)} \over {2\sin 2\alpha \sin 2\beta }}} \right]$$

$$ = {{16} \over 3}\left[ {{{\sin (\alpha + \beta )\cos (\alpha - \beta )} \over {\cos (2\alpha - 2\beta ) - cos(2\alpha + 2\beta )}}} \right]$$

$$ = {{16} \over 3}\left[ {{{{1 \over 3} \times {2 \over 3}} \over {(2{{\cos }^2}(\alpha - \beta ) - 1) - (1 - 2{{\sin }^2}(\alpha + \beta ))}}} \right]$$

$$ = {{32} \over {27}}\left[ {{1 \over {2 \times {4 \over 9} - 2 + 2 \times {1 \over 9}}}} \right]$$

$$ = {{32} \over {27}}\left[ {{9 \over {8 - 18 + 2}}} \right]$$

$$ = {{32} \over {27}}\left[ {{9 \over { - 8}}} \right]$$

$$ = - {4 \over 3}$$

$$\therefore$$ $${E^2} = {{16} \over 9} = 1.77$$

$$ \Rightarrow [{E^2}] = 1$$