If `x + y + z = xyz and x, y, z gt 0`, then find the value of `tan^(-1) x + tan^(-1) y + tan^(-1) z`

Explanation

To solve the problem, we need to find the value of \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \) given that \( x + y + z = xyz \) and \( x, y, z > 0 \).

### Step-by-Step Solution:

1. **Substituting Variables**:
Let's set \( x = \tan A \), \( y = \tan B \), and \( z = \tan C \) for some angles \( A, B, C \). This means we will express \( \tan^{-1} x \), \( \tan^{-1} y \), and \( \tan^{-1} z \) in terms of these angles:
\[ \tan^{-1} x = A, \quad \tan^{-1} y = B, \quad \tan^{-1} z = C \]

2. **Using the Given Condition**:
The condition \( x + y + z = xyz \) translates to:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]

3. **Using the Tangent Addition Formula**:
We know from the tangent addition formula that:
\[ \tan(A + B + C) = \frac{\tan A + \tan B + \tan C}{1 - \tan A \tan B \tan C} \]
Since \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \), we can substitute this into the formula:
\[ \tan(A + B + C) = \frac{\tan A \tan B \tan C}{1 - \tan A \tan B \tan C} \]

4. **Setting Up the Equation**:
From the equation above, we can see that if we set \( A + B + C = \frac{\pi}{k} \) for some integer \( k \), then:
\[ \tan(A + B + C) = \tan\left(\frac{\pi}{k}\right) \]
This means that:
\[ A + B + C = n\pi \quad \text{for some integer } n \]

5. **Conclusion**:
Given that \( x, y, z > 0 \), we can conclude that \( A + B + C = \frac{\pi}{k} \) for \( k = 1 \) (the simplest case). Therefore:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]

### Final Answer:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]