For `x, y, z, t in R, sin^(-1) x + cos^(-1) y + sec^(-1) z ge t^(2) - sqrt(2pi t) + 3pi`
The value of `x + y + z` is equal to

Options

A. 1
B. 0
C. 2
D. `-1`

Explanation

To solve the problem, we need to analyze the given inequality involving inverse trigonometric functions and find the values of \(x\), \(y\), and \(z\) such that we can compute \(x + y + z\).

### Step-by-Step Solution:

1. **Understand the ranges of the inverse trigonometric functions:**
- The range of \(\sin^{-1} x\) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- The range of \(\cos^{-1} y\) is \([0, \pi]\).
- The range of \(\sec^{-1} z\) is \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\).

2. **Determine the maximum values of the functions:**
- The maximum value of \(\sin^{-1} x\) is \(\frac{\pi}{2}\).
- The maximum value of \(\cos^{-1} y\) is \(\pi\).
- The maximum value of \(\sec^{-1} z\) is \(\pi\).

3. **Combine the maximum values:**
\[ \sin^{-1} x + \cos^{-1} y + \sec^{-1} z \leq \frac{\pi}{2} + \pi + \pi = \frac{5\pi}{2} \]

4. **Set up the inequality:**
We have:
\[ \sin^{-1} x + \cos^{-1} y + \sec^{-1} z \geq t^2 - \sqrt{2\pi t} + 3\pi \]
Therefore, we can write:
\[ t^2 - \sqrt{2\pi t} + 3\pi \leq \frac{5\pi}{2} \]

5. **Rearranging the inequality:**
\[ t^2 - \sqrt{2\pi t} + 3\pi - \frac{5\pi}{2} \leq 0 \]
Simplifying gives:
\[ t^2 - \sqrt{2\pi t} + \frac{1\pi}{2} \leq 0 \]

6. **Complete the square:**
To analyze the quadratic, we can complete the square:
\[ 2(t - \frac{\pi}{4})^2 + \frac{1\pi}{2} - \frac{\pi}{2} \leq 0 \]
This implies:
\[ (t - \frac{\pi}{4})^2 \geq 0 \]
Thus, \(t\) must be at least \(\frac{\pi}{2}\).

7. **Finding values of \(x\), \(y\), and \(z\):**
- Set \(\sin^{-1} x = \frac{\pi}{2} \Rightarrow x = 1\).
- Set \(\cos^{-1} y = \pi \Rightarrow y = -1\).
- Set \(\sec^{-1} z = \pi \Rightarrow z = -1\).

8. **Calculate \(x + y + z\):**
\[ x + y + z = 1 + (-1) + (-1) = 1 - 1 - 1 = -1 \]

### Final Answer:
The value of \(x + y + z\) is \(-1\).