If `tanA-tan B=x`, and `cot B-cotA=y`, then find the value of `cot(A-B)`.

Explanation

To find the value of \( \cot(A - B) \) given that \( \tan A - \tan B = x \) and \( \cot B - \cot A = y \), we can follow these steps:

### Step 1: Use the cotangent subtraction formula
The formula for \( \cot(A - B) \) is given by:
\[ \cot(A - B) = \frac{1}{\tan(A - B)} \]
We will first find \( \tan(A - B) \).

### Step 2: Apply the tangent subtraction formula
The tangent subtraction formula states:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
From the problem, we know \( \tan A - \tan B = x \). Thus, we can rewrite it as:
\[ \tan(A - B) = \frac{x}{1 + \tan A \tan B} \]

### Step 3: Express \( \tan A \tan B \) in terms of \( y \)
We also know that:
\[ \cot B - \cot A = y \]
We can express cotangent in terms of tangent:
\[ \cot B = \frac{1}{\tan B}, \quad \cot A = \frac{1}{\tan A} \]
Thus:
\[ \frac{1}{\tan B} - \frac{1}{\tan A} = y \]
This can be rewritten as:
\[ \frac{\tan A - \tan B}{\tan A \tan B} = y \]
Substituting \( \tan A - \tan B = x \):
\[ \frac{x}{\tan A \tan B} = y \]
From this, we can express \( \tan A \tan B \):
\[ \tan A \tan B = \frac{x}{y} \]

### Step 4: Substitute back into the tangent subtraction formula
Now we can substitute \( \tan A \tan B \) back into the formula for \( \tan(A - B) \):
\[ \tan(A - B) = \frac{x}{1 + \frac{x}{y}} = \frac{x}{\frac{y + x}{y}} = \frac{xy}{y + x} \]

### Step 5: Find \( \cot(A - B) \)
Now we can find \( \cot(A - B) \):
\[ \cot(A - B) = \frac{1}{\tan(A - B)} = \frac{y + x}{xy} \]

### Final Answer
Thus, the value of \( \cot(A - B) \) is:
\[ \cot(A - B) = \frac{y + x}{xy} \]