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We need to find the angle whose secant is $-\sqrt{2}$. Since the range of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$, we are looking for an angle in the second quadrant where secant is negative. $\sec(\frac{3\pi}{4}) = -\sqrt{2}$, so $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$.
We need to find the angle whose tangent is $\frac{1}{\sqrt{3}}$. Since the range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, we are looking for an angle in the first quadrant where tangent is positive. $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, so $\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
Now, we subtract the two values: $\sec^{-1}(-\sqrt{2}) - \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{3\pi}{4} - \frac{\pi}{6}$
To subtract the fractions, we need a common denominator, which is 12. $\frac{3\pi}{4} - \frac{\pi}{6} = \frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12}$
Final Answer: \(\frac{7\pi}{12}\)