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

Options

A. `(pi)`
B. `pi//2`
C. 1
D. none of these

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 formula for the sum of inverse tangents**:
We know that:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Applying this to \( \tan^{-1}x + \tan^{-1}y \):
\[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]

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 sum formula again:
\[ = \tan^{-1}\left(\frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z}\right) \]

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

4. **Substituting \( xy + yz + zx = 1 \)**:
We know from the problem statement that \( xy + yz + zx = 1 \). Thus, we can substitute \( yz + zx = 1 - xy \) into our expression.

5. **Final expression**:
The denominator simplifies to:
\[ 1 - xy - z(x + y) = 1 - xy - (1 - xy) = 0 \]
This indicates that the expression approaches infinity.

6. **Conclusion**:
Since the denominator approaches zero, the overall expression for \( \tan^{-1}(x + y + z) \) becomes:
\[ \tan^{-1}\left(\frac{\text{something}}{0}\right) \rightarrow \frac{\pi}{2} \]

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