We need to evaluate the expression $\sqrt{11 - 2\sqrt{30}} - \frac{1}{\sqrt{11 - 2\sqrt{30}}}$.
To simplify $\sqrt{11 - 2\sqrt{30}}$, we look for two numbers whose sum is 11 and product is 30. These numbers are 6 and 5.
Therefore, we can rewrite the expression under the square root as:
$11 - 2\sqrt{30} = 6 + 5 - 2\sqrt{6 \times 5} = (\sqrt{6})^2 + (\sqrt{5})^2 - 2\sqrt{6}\sqrt{5} = (\sqrt{6} - \sqrt{5})^2$
So, $\sqrt{11 - 2\sqrt{30}} = \sqrt{(\sqrt{6} - \sqrt{5})^2}$. Since $\sqrt{6} > \sqrt{5}$, the absolute value is $\sqrt{6} - \sqrt{5}$.
Let $x = \sqrt{11 - 2\sqrt{30}} = \sqrt{6} - \sqrt{5}$.
Now, we find the reciprocal of $x$:
$ \frac{1}{x} = \frac{1}{\sqrt{6} - \sqrt{5}} $
Rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{6} + \sqrt{5})$:
$ \frac{1}{x} = \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = \frac{\sqrt{6} + \sqrt{5}}{1} = \sqrt{6} + \sqrt{5} $
Substitute the values of $x$ and $\frac{1}{x}$ back into the original expression form:
$ x - \frac{1}{x} = (\sqrt{6} - \sqrt{5}) - (\sqrt{6} + \sqrt{5}) $
$ = \sqrt{6} - \sqrt{5} - \sqrt{6} - \sqrt{5} $
$ = (\sqrt{6} - \sqrt{6}) + (-\sqrt{5} - \sqrt{5}) $
$ = 0 - 2\sqrt{5} $
$ = -2\sqrt{5} $
The value of the expression $\sqrt{11 - 2\sqrt{30}} - \frac{1}{\sqrt{11 - 2\sqrt{30}}}$ is $-2\sqrt{5}$.
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