If tan A - tan B =x and cot B - cot A=y, then: cot(A-B)=

Options

A. `1/x -1/y`
B. `1/x + 1/y`
C. `1/y-1/x`
D. `x-y`

Answer

To solve the problem, we need to find the expression for \( \cot(A - B) \) given that \( \tan A - \tan B = x \) and \( \cot B - \cot A = y \).

### Step-by-Step Solution:

1. **Use the formula for \( \tan(A - B) \)**:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Given \( \tan A - \tan B = x \), we can substitute this into the formula:
\[ \tan(A - B) = \frac{x}{1 + \tan A \tan B} \]

2. **Express \( \cot(A - B) \)**:
Since \( \cot(A - B) = \frac{1}{\tan(A - B)} \), we can write:
\[ \cot(A - B) = \frac{1 + \tan A \tan B}{x} \]

3. **Use the information about \( \cot A \) and \( \cot B \)**:
We know that:
\[ \cot B - \cot A = y \]
This can be rewritten using the definitions of cotangent:
\[ \frac{1}{\tan B} - \frac{1}{\tan A} = y \]
Taking the common denominator:
\[ \frac{\tan A - \tan B}{\tan A \tan B} = y \]
Substituting \( \tan A - \tan B = x \):
\[ \frac{x}{\tan A \tan B} = y \]
From this, we can express \( \tan A \tan B \):
\[ \tan A \tan B = \frac{x}{y} \]

4. **Substitute back into the expression for \( \cot(A - B) \)**:
Now, substituting \( \tan A \tan B = \frac{x}{y} \) into the expression for \( \cot(A - B) \):
\[ \cot(A - B) = \frac{1 + \frac{x}{y}}{x} \]
Simplifying this:
\[ \cot(A - B) = \frac{y + x}{xy} \]

5. **Final Result**:
Therefore, the final expression for \( \cot(A - B) \) is:
\[ \cot(A - B) = \frac{y + x}{xy} \]