If xy +yz+zx=1 then `tan^(-1)x+tan^(-1)y+tan^(-1)z`=

Answer

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

### Step-by-Step Solution

1. **Use the Identity for Sum of Inverse Tangents**:
We start with the identity:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
We will apply this identity to the first two terms, \( \tan^{-1} x \) and \( \tan^{-1} y \):
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]

2. **Add \( \tan^{-1} z \)**:
Now, we add \( \tan^{-1} z \) to the result:
\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z \]
Using the identity again:
\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z = \tan^{-1} \left( \frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z} \right) \]

3. **Simplify the Numerator**:
The numerator becomes:
\[ \frac{x + y + z(1 - xy)}{1 - xy - z(x + y)} \]
This simplifies to:
\[ x + y + z - xyz \]

4. **Simplify the Denominator**:
The denominator becomes:
\[ 1 - xy - zx - zy \]
By substituting \( xy + yz + zx = 1 \), we have:
\[ 1 - 1 = 0 \]

5. **Final Expression**:
Therefore, we have:
\[ \tan^{-1} \left( \frac{x + y + z - xyz}{0} \right) \]
Since division by zero leads to infinity, we conclude:
\[ \tan^{-1}(\infty) = \frac{\pi}{2} \]

### Conclusion
Thus, we find that:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]