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
We know that:
\[ \cot(A - B) = \frac{1}{\tan(A - B)} \]
And the formula for \( \tan(A - B) \) is:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

### Step 2: Substitute the values into the formula
Substituting the expression for \( \tan(A - B) \) into the cotangent formula gives:
\[ \cot(A - B) = \frac{1 + \tan A \tan B}{\tan A - \tan B} \]
Since \( \tan A - \tan B = x \), we can rewrite this as:
\[ \cot(A - B) = \frac{1 + \tan A \tan B}{x} \]

### Step 3: Express \( \tan A \tan B \) in terms of \( y \)
From the second equation, \( \cot B - \cot A = y \), we can express \( \cot A \) and \( \cot B \) in terms of \( \tan A \) and \( \tan B \):
\[ \cot A = \frac{1}{\tan A}, \quad \cot B = \frac{1}{\tan B} \]
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 cotangent expression
Now we substitute \( \tan A \tan B = \frac{x}{y} \) back into the expression for \( \cot(A - B) \):
\[ \cot(A - B) = \frac{1 + \frac{x}{y}}{x} \]
This simplifies to:
\[ \cot(A - B) = \frac{y + x}{xy} \]

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