If `tanA-tanB=x and cotB-cotA=y`, then `cot(A-B)=?`

Options

A. `(1)/(x)+y`
B. `(1)/(xy)`
C. `(1)/(x)-(1)/(y)`
D. `(1)/(x)+(1)/(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. **Start with the given equations**:
\[ \tan A - \tan B = x \]
\[ \cot B - \cot A = y \]

2. **Express \( \tan A \) and \( \tan B \) in terms of \( \cot A \) and \( \cot B \)**:
We know that \( \tan A = \frac{1}{\cot A} \) and \( \tan B = \frac{1}{\cot B} \). Thus, we can rewrite the first equation:
\[ \frac{1}{\cot A} - \frac{1}{\cot B} = x \]

3. **Find a common denominator**:
The common denominator for the left-hand side is \( \cot A \cot B \):
\[ \frac{\cot B - \cot A}{\cot A \cot B} = x \]

4. **Substituting \( \cot B - \cot A \)**:
From the second equation, we have \( \cot B - \cot A = y \). Substitute this into the equation:
\[ \frac{y}{\cot A \cot B} = x \]

5. **Rearranging the equation**:
Rearranging gives us:
\[ \cot A \cot B = \frac{y}{x} \]

6. **Using the formula for \( \cot(A - B) \)**:
The formula for \( \cot(A - B) \) is:
\[ \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \]
Substitute \( \cot B - \cot A = -y \) (since \( \cot A - \cot B = -y \)):
\[ \cot(A - B) = \frac{\cot A \cot B + 1}{-y} \]

7. **Substituting \( \cot A \cot B \)**:
We already found that \( \cot A \cot B = \frac{y}{x} \):
\[ \cot(A - B) = \frac{\frac{y}{x} + 1}{-y} \]

8. **Simplifying the expression**:
Combine the terms in the numerator:
\[ \cot(A - B) = \frac{\frac{y + x}{x}}{-y} = -\frac{y + x}{xy} \]

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

### Final Answer:
\[ \cot(A - B) = -\frac{y + x}{xy} \]