Step 1: Understanding the Question:
The question asks to find the sum of principal values of three inverse trigonometric functions.
Step 2: Key Formula or Approach:
The principal value ranges for the functions are:
- $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
- $\sec^{-1}(x) \in [0, \pi] - \left\{\frac{\pi}{2}\right\}$
- $\sin^{-1}(x) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Step 3: Detailed Explanation:
1. For $\tan^{-1}(\sqrt{3})$:
We know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.
So, $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
2. For $\sec^{-1}(-2)$:
Let $\sec^{-1}(-2) = \theta \implies \sec \theta = -2$.
Since the range is $[0, \pi]$, we have $\sec \theta = -\sec\left(\frac{\pi}{3}\right) = \sec\left(\pi - \frac{\pi}{3}\right) = \sec\left(\frac{2\pi}{3}\right)$.
So, $\sec^{-1}(-2) = \frac{2\pi}{3}$.
3. For $\sin^{-1}\left(-\frac{1}{2}\right)$:
We know $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, and since $\sin^{-1}(-x) = -\sin^{-1}(x)$,
$\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.
Combining the values:
\[ \text{Value} = \frac{\pi}{3} + \frac{2\pi}{3} - \left(-\frac{\pi}{6}\right) \]
\[ \text{Value} = \pi + \frac{\pi}{6} \]
\[ \text{Value} = \frac{7\pi}{6} \]
Step 4: Final Answer:
The calculated value is $\frac{7\pi}{6}$.