If `x+y+z=xyz`, 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 where \( x + y + z = xyz \) and we need to find \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Rearranging the Equation**: Rearranging gives:
\[ x + y = xyz - z \]
We can factor out \( z \) from the right side:
\[ x + y = z(xy - 1) \]

3. **Using Inverse Tangent**: Let:
\[ \tan^{-1}x = \alpha, \quad \tan^{-1}y = \beta \]
Then:
\[ x = \tan \alpha, \quad y = \tan \beta \]

4. **Applying the Tangent Addition Formula**: We can use the identity for the tangent of a sum:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting \( x \) and \( y \):
\[ \tan(\alpha + \beta) = \frac{x + y}{1 - xy} \]

5. **Substituting for \( x + y \)**: From step 2, substitute \( x + y \):
\[ \tan(\alpha + \beta) = \frac{z(xy - 1)}{1 - xy} \]

6. **Simplifying the Expression**: This can be rewritten as:
\[ \tan(\alpha + \beta) = -z \]
(since \( z(xy - 1) \) can be rewritten as \( -z(1 - xy) \)).

7. **Finding the Angle**: Therefore, we have:
\[ \alpha + \beta = \tan^{-1}(-z) \]

8. **Final Expression**: Now, adding \( \tan^{-1}z \) to both sides:
\[ \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}(-z) + \tan^{-1}z \]

9. **Using the Property of Inverse Tangents**: We know that:
\[ \tan^{-1}(-z) + \tan^{-1}z = 0 \]

10. **Conclusion**: Thus, we conclude that:
\[ \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = 0 \]

### Final Answer:
\[ \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = 0 \]

---