If `x gt y gt z gt 0`, then find the value of `"cot"^(-1) (xy + 1)/(x - y) + "cot"^(-1)(yz + 1)/(y - z) + "cot"^(-1)(zx + 1)/(z - x)`

Options

A. `pi`
B. `pi/2`
C. `pi/3`
D. `pi/4`

Explanation

To solve the given problem step by step, we need to evaluate the expression:

\[ \cot^{-1} \left( \frac{xy + 1}{x - y} \right) + \cot^{-1} \left( \frac{yz + 1}{y - z} \right) + \cot^{-1} \left( \frac{zx + 1}{z - x} \right) \]

### Step 1: Rewrite each term using cotangent identity
Using the identity for the cotangent of the difference of two angles, we can express each term in the form of an inverse tangent function. The identity states that:

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

Thus, we can rewrite each term as follows:

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

Similarly, we can rewrite the other two terms:

\[ \cot^{-1} \left( \frac{yz + 1}{y - z} \right) = \tan^{-1} \left( \frac{y - z}{yz + 1} \right) \]

\[ \cot^{-1} \left( \frac{zx + 1}{z - x} \right) = \tan^{-1} \left( \frac{z - x}{zx + 1} \right) \]

### Step 2: Combine the terms
Now we can combine these terms:

\[ \tan^{-1} \left( \frac{x - y}{xy + 1} \right) + \tan^{-1} \left( \frac{y - z}{yz + 1} \right) + \tan^{-1} \left( \frac{z - x}{zx + 1} \right) \]

### Step 3: Use the tangent addition formula
We can use the tangent addition formula:

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

We will apply this formula iteratively to combine the terms.

### Step 4: Apply the addition formula
Let:

\[ A = \tan^{-1} \left( \frac{x - y}{xy + 1} \right) \]
\[ B = \tan^{-1} \left( \frac{y - z}{yz + 1} \right) \]
\[ C = \tan^{-1} \left( \frac{z - x}{zx + 1} \right) \]

First, combine \(A\) and \(B\):

\[ A + B = \tan^{-1} \left( \frac{\frac{x - y}{xy + 1} + \frac{y - z}{yz + 1}}{1 - \frac{(x - y)(y - z)}{(xy + 1)(yz + 1)}} \right) \]

Then, add \(C\) to the result.

### Step 5: Simplify the expression
After combining all three angles, we notice that the terms will cancel out due to the cyclic nature of the variables \(x\), \(y\), and \(z\).

### Step 6: Conclusion
The final result simplifies to:

\[ A + B + C = 0 \]

Thus, the value of the original expression is:

\[ \boxed{0} \]