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 1: Rewrite the equation
Given the equation:
\[ x + y + z = xyz \]
We can rearrange it as:
\[ x + y = xyz - z \]

### Step 2: Define angles
Let:
\[ \tan^{-1} x = \alpha, \quad \tan^{-1} y = \beta, \quad \tan^{-1} z = \gamma \]
Thus, we have:
\[ x = \tan \alpha, \quad y = \tan \beta, \quad z = \tan \gamma \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula, we know:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting \( x \) and \( y \) gives:
\[ \tan(\alpha + \beta) = \frac{x + y}{1 - xy} \]

### Step 4: Substitute \( x + y \)
From our rearranged equation \( x + y = xyz - z \), we substitute this into the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{xyz - z}{1 - xy} \]

### Step 5: Relate to \( z \)
Now, we can express \( \tan(\alpha + \beta + \gamma) \):
\[ \tan(\alpha + \beta + \gamma) = \frac{\tan(\alpha + \beta) + \tan \gamma}{1 - \tan(\alpha + \beta) \tan \gamma} \]
Substituting \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta + \gamma) = \frac{\frac{xyz - z}{1 - xy} + z}{1 - \frac{(xyz - z)z}{1 - xy}} \]

### Step 6: Simplify the expression
To simplify this expression, we can analyze the condition \( x + y + z = xyz \). This implies that:
\[ \tan(\alpha + \beta + \gamma) = 0 \]
which means:
\[ \alpha + \beta + \gamma = n\pi \quad \text{for some integer } n \]

### Step 7: Conclusion
Since we are looking for \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z \), we conclude:
\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]