If `sin^(-1)x+sin^(-1)y+sin^(-1)z=(3pi)/(2),` then `A=x^(2)+y^(2)+z^(2)`.
`(sin^(-1)x)^(2)+(sin^(-1)y)^(2)+(sin^(-1)z)^(2)=(3pi^(2))/(4)`, then `B=|(x+y+z)_("min")|`. Then the value of `(A+B)` is _____________ .

Answer

To solve the problem, we need to analyze the given equations step by step.

### Step 1: Analyze the first equation
We are given that:
\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2} \]
The range of \(\sin^{-1}\) function is \([- \frac{\pi}{2}, \frac{\pi}{2}]\). The only way for the sum of three angles (each in this range) to equal \(\frac{3\pi}{2}\) is if each angle is equal to \(\frac{\pi}{2}\). Therefore:
\[ \sin^{-1} x = \frac{\pi}{2}, \quad \sin^{-1} y = \frac{\pi}{2}, \quad \sin^{-1} z = \frac{\pi}{2} \]
This implies:
\[ x = 1, \quad y = 1, \quad z = 1 \]

### Step 2: Calculate \(A\)
Now we can calculate \(A\):
\[ A = x^2 + y^2 + z^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3 \]

### Step 3: Analyze the second equation
Next, we look at the second equation:
\[ (\sin^{-1} x)^2 + (\sin^{-1} y)^2 + (\sin^{-1} z)^2 = \frac{3\pi^2}{4} \]
Substituting the values we found:
\[ \left(\frac{\pi}{2}\right)^2 + \left(\frac{\pi}{2}\right)^2 + \left(\frac{\pi}{2}\right)^2 = \frac{3\pi^2}{4} \]
This confirms that our values of \(x\), \(y\), and \(z\) are correct.

### Step 4: Calculate \(B\)
Now we need to find \(B\):
\[ B = |(x+y+z)_{min}| \]
Since \(x = 1\), \(y = 1\), and \(z = 1\), the minimum value of \(x + y + z\) occurs when we consider the possible minimum values of \(x\), \(y\), and \(z\) in the context of the sine inverse function. The minimum values of \(x\), \(y\), and \(z\) can be \(-1\) (since \(\sin^{-1}(-1) = -\frac{\pi}{2}\)). Thus:
\[ x + y + z = -1 - 1 - 1 = -3 \]
Therefore:
\[ B = |-3| = 3 \]

### Step 5: Calculate \(A + B\)
Finally, we calculate \(A + B\):
\[ A + B = 3 + 3 = 6 \]

Thus, the final answer is:
\[ \boxed{6} \]