For `x,y,z,t in R`, if `sin^(-1)x+cos^(-1)y+sec^(-1)z ge t^(2)-sqrt(2pi)*t+3pi`, then the value of `tan^(-1)x+tan^(-1)y+tan^(-1)z+tan^(-1)( sqrt((2)/(pi))t)` is __________.

Answer

To solve the problem step by step, we start with the given inequality:

\[ \sin^{-1} x + \cos^{-1} y + \sec^{-1} z \geq t^2 - \sqrt{2\pi} t + 3\pi \]

### Step 1: Determine 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 \([0, \frac{\pi}{2}] \cup [\frac{\pi}{2}, \pi]\).

### Step 2: Calculate the maximum value of the left-hand side
Adding the maximum values from each range:
\[ \sin^{-1} x + \cos^{-1} y + \sec^{-1} z \leq \frac{\pi}{2} + \pi + \pi = \frac{5\pi}{2} \]

### Step 3: Set up the inequality
Now we have:
\[ \frac{5\pi}{2} \geq t^2 - \sqrt{2\pi} t + 3\pi \]

### Step 4: Rearranging the inequality
Rearranging gives:
\[ t^2 - \sqrt{2\pi} t + 3\pi - \frac{5\pi}{2} \leq 0 \]
This simplifies to:
\[ t^2 - \sqrt{2\pi} t + \frac{1\pi}{2} \leq 0 \]

### Step 5: Completing the square
To complete the square:
\[ t^2 - \sqrt{2\pi} t + \left(\frac{\sqrt{2\pi}}{2}\right)^2 - \left(\frac{\sqrt{2\pi}}{2}\right)^2 + \frac{\pi}{2} \leq 0 \]
This becomes:
\[ \left(t - \frac{\sqrt{2\pi}}{2}\right)^2 - \frac{2\pi}{4} + \frac{\pi}{2} \leq 0 \]
Simplifying:
\[ \left(t - \frac{\sqrt{2\pi}}{2}\right)^2 \leq 0 \]
This implies:
\[ t - \frac{\sqrt{2\pi}}{2} = 0 \implies t = \frac{\sqrt{2\pi}}{2} \]

### Step 6: Finding values of \(x\), \(y\), and \(z\)
Substituting \(t = \frac{\sqrt{2\pi}}{2}\):
- Since \(\sin^{-1} x = \frac{\pi}{2}\), we have \(x = 1\).
- Since \(\cos^{-1} y = \pi\), we have \(y = -1\).
- Since \(\sec^{-1} z = \pi\), we have \(z = -1\).

### Step 7: Calculate \(tan^{-1} x + tan^{-1} y + tan^{-1} z + tan^{-1}(\sqrt{\frac{2}{\pi}} t)\)
Now substituting \(x\), \(y\), and \(z\):
\[ \tan^{-1} 1 + \tan^{-1} (-1) + \tan^{-1} (-1) + \tan^{-1} \left(\sqrt{\frac{2}{\pi}} \cdot \frac{\sqrt{2\pi}}{2}\right) \]
Calculating:
- \(\tan^{-1} 1 = \frac{\pi}{4}\)
- \(\tan^{-1} (-1) = -\frac{\pi}{4}\)
- \(\tan^{-1} (-1) = -\frac{\pi}{4}\)
- \(\tan^{-1} \left(\sqrt{\frac{2}{\pi}} \cdot \frac{\sqrt{2\pi}}{2}\right) = \tan^{-1} 1 = \frac{\pi}{4}\)

### Step 8: Final calculation
Combining these:
\[ \frac{\pi}{4} - \frac{\pi}{4} - \frac{\pi}{4} + \frac{\pi}{4} = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]