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

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ x + y + z = xyz \]

2. **Rearrange the equation**:
We can rearrange the equation to express \( z \):
\[ z = \frac{x + y}{xy - 1} \]
(This step is derived from rearranging the original equation.)

3. **Apply the tangent addition formula**:
We know that:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]
if \( xy < 1 \).

4. **Consider the case when \( xy < 1 \)**:
In this case, we can write:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]
Now, we can add \( \tan^{-1} z \):
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z \]

5. **Use the tangent addition formula again**:
Using the formula again:
\[ \tan^{-1} \left( \frac{a + b}{1 - ab} \right) + \tan^{-1} c = \tan^{-1} \left( \frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z} \right) \]
Substitute \( z = \frac{x + y}{xy - 1} \):
\[ \tan^{-1} \left( \frac{x + y + z(1 - xy)}{1 - z \cdot \frac{x + y}{1 - xy}} \right) \]

6. **Simplify the expression**:
After substituting and simplifying, we find:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]

7. **Conclusion**:
Therefore, the value of \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \) is:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]