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

Options

A. 0
B. 1
C. `1/(xyz)`
D. xyz

Answer

To solve the equation \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi \) and find the value of \( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \), we can follow these steps:

### Step 1: Use the formula for the sum of inverse tangents
We start with the equation:
\[ \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi \]
Using the formula for the sum of three inverse tangents:
\[ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c = \tan^{-1}\left(\frac{a + b + c - abc}{1 - (ab + ac + bc)}\right) \]
we can rewrite our equation as:
\[ \tan^{-1}\left(x + y + z - xyz\right) = \pi \]

### Step 2: Apply the tangent function
Taking the tangent of both sides, we have:
\[ x + y + z - xyz = \tan(\pi) \]
Since \( \tan(\pi) = 0 \), we get:
\[ x + y + z - xyz = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ x + y + z = xyz \]

### Step 4: Express the desired quantity
We want to find:
\[ \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \]
This can be rewritten as:
\[ \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = \frac{z + x + y}{xyz} \]

### Step 5: Substitute the value from Step 3
From Step 3, we know that \( x + y + z = xyz \). Substituting this into our expression gives:
\[ \frac{z + x + y}{xyz} = \frac{xyz}{xyz} = 1 \]

### Conclusion
Thus, the value of \( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \) is:
\[ \boxed{1} \]