If `sin^(-1)x+sin^(-1)y+sin^(-1)z=(pi)/(2)`, then the value of `x^(2)+y^(2)+z^(2)+2xyz` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 + z^2 + 2xyz \) given that \( \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Let \( \sin^{-1}x = \alpha \), \( \sin^{-1}y = \beta \), and \( \sin^{-1}z = \gamma \)**:
\[ \alpha + \beta + \gamma = \frac{\pi}{2} \]

2. **Rearranging the equation**:
\[ \alpha + \beta = \frac{\pi}{2} - \gamma \]

3. **Taking the cosine of both sides**:
\[ \cos(\alpha + \beta) = \cos\left(\frac{\pi}{2} - \gamma\right) \]

4. **Using the identity \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta \)**:
\[ \cos(\alpha + \beta) = \sin\gamma \]

5. **Applying the cosine addition formula**:
\[ \cos\alpha \cos\beta - \sin\alpha \sin\beta = \sin\gamma \]

6. **Substituting \( \sin\alpha = x \), \( \sin\beta = y \), and \( \sin\gamma = z \)**:
\[ \cos\alpha \cos\beta - xy = z \]

7. **Using the identity \( \cos^2\theta = 1 - \sin^2\theta \)**:
\[ \cos\alpha = \sqrt{1 - x^2}, \quad \cos\beta = \sqrt{1 - y^2} \]

8. **Substituting these into the equation**:
\[ \sqrt{1 - x^2} \sqrt{1 - y^2} - xy = z \]

9. **Squaring both sides**:
\[ (1 - x^2)(1 - y^2) = (z + xy)^2 \]

10. **Expanding both sides**:
\[ 1 - x^2 - y^2 + x^2y^2 = z^2 + 2xyz + x^2y^2 \]

11. **Simplifying the equation**:
\[ 1 - x^2 - y^2 = z^2 + 2xyz \]

12. **Rearranging gives**:
\[ x^2 + y^2 + z^2 + 2xyz = 1 \]

### Final Answer:
\[ x^2 + y^2 + z^2 + 2xyz = 1 \]