If `(sin^(-1)x)^2+(sin^(-1)y)^2+(sin^(-1)z)^2=3/4pi^2` , find the value of `x^2+y^2+z^2` .

Explanation

To solve the problem, we need to find the value of \( x^2 + y^2 + z^2 \) given that

\[ (\sin^{-1} x)^2 + (\sin^{-1} y)^2 + (\sin^{-1} z)^2 = \frac{3}{4} \pi^2. \]

### Step-by-Step Solution:

1. **Understanding the Range of \( \sin^{-1} x \)**:
The function \( \sin^{-1} x \) (inverse sine) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) for \( x \) in the interval \([-1, 1]\). Therefore, the maximum value of \( \sin^{-1} x \) is \( \frac{\pi}{2} \).

2. **Finding the Maximum Value of the Squares**:
Since \( \sin^{-1} x \) can take values from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), the maximum value of \( (\sin^{-1} x)^2 \) is:

\[ \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}. \]

Thus, we have:

\[ (\sin^{-1} x)^2 \leq \frac{\pi^2}{4}, \quad (\sin^{-1} y)^2 \leq \frac{\pi^2}{4}, \quad (\sin^{-1} z)^2 \leq \frac{\pi^2}{4}. \]

3. **Adding the Inequalities**:
Adding these inequalities gives:

\[ (\sin^{-1} x)^2 + (\sin^{-1} y)^2 + (\sin^{-1} z)^2 \leq 3 \cdot \frac{\pi^2}{4} = \frac{3\pi^2}{4}. \]

This matches the condition given in the problem:

\[ (\sin^{-1} x)^2 + (\sin^{-1} y)^2 + (\sin^{-1} z)^2 = \frac{3}{4} \pi^2. \]

4. **Equality Condition**:
The equality holds when each term reaches its maximum value. Therefore, we can conclude:

\[ \sin^{-1} x = \pm \frac{\pi}{2}, \quad \sin^{-1} y = \pm \frac{\pi}{2}, \quad \sin^{-1} z = \pm \frac{\pi}{2}. \]

5. **Finding Values of \( x, y, z \)**:
From \( \sin^{-1} x = \pm \frac{\pi}{2} \), we have:

\[ x = \pm 1, \quad y = \pm 1, \quad z = \pm 1. \]

6. **Calculating \( x^2 + y^2 + z^2 \)**:
Now, we calculate \( x^2 + y^2 + z^2 \):

\[ x^2 + y^2 + z^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3. \]

### Final Answer:
Thus, the value of \( x^2 + y^2 + z^2 \) is

\[ \boxed{3}. \]