If `cot^(-1)x+cot^(-1)y+cot^(-1)z=pi/2,x , y , z >0a n dx y<1,` then `x+y+z` is also equal to `1/x+1/y+1/z` (b) `x y z` `x y+y z+z x` (d) none of these

Options

A. `(1)/(x) + (1)/(y) + (1)/(z)`
B. `xyz`
C. `xy + yz + zx`
D. none of these

Explanation

To solve the problem, we start with the given equation:

\[ \cot^{-1} x + \cot^{-1} y + \cot^{-1} z = \frac{\pi}{2} \]

### Step 1: Use the identity for cotangent
We know that:

\[ \cot^{-1} a + \cot^{-1} b = \cot^{-1} \left( \frac{ab - 1}{a + b} \right) \]

This means we can rewrite the sum of the inverse cotangents in terms of a single inverse cotangent.

### Step 2: Rewrite the equation
Using the identity, we can express the equation as:

\[ \cot^{-1} x + \cot^{-1} y = \frac{\pi}{2} - \cot^{-1} z \]

### Step 3: Convert to tangent
Taking the cotangent of both sides gives us:

\[ \frac{xy - 1}{x + y} = \tan\left(\frac{\pi}{2} - \cot^{-1} z\right) = \frac{1}{z} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:

\[ (z)(xy - 1) = (x + y) \]

### Step 5: Rearranging the equation
Rearranging the equation leads to:

\[ zxy - z = x + y \]

### Step 6: Isolate terms
We can rearrange this to isolate \(x + y + z\):

\[ x + y + z = zxy \]

### Conclusion
Thus, we have shown that:

\[ x + y + z = xyz \]

The answer is option (b) \(xyz\).