Step 1: Value of tan inverse root 3:
Using the standard result for the principal branch, $\tan^{-1}\sqrt3 = \dfrac{\pi}{3}$, since $\tan(\pi/3) = \sqrt3$ and $\pi/3$ lies in the principal range.
Step 2: Use the identity for sec inverse of a negative number:
For $x \ge 1$, the identity states $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$.
Here $x = 2$, and since $\sec(\pi/3) = 2$, we have $\sec^{-1}(2) = \dfrac{\pi}{3}$.
Applying the identity:
\[ \sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
Step 3: Combine the results:
Subtract the two computed values as required by the expression:
\[ \tan^{-1}\sqrt3 - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \]
Final Answer:
Using the negative-argument identity for sec inverse gives the same result.
\[ \boxed{-\frac{\pi}{3}} \]