If `tantheta-tanphi=xandcotphi-cottheta=y`, then `cot(theta-phi)=?`

Options

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

Answer

To solve the problem, we need to find \( \cot(\theta - \phi) \) given the equations \( \tan \theta - \tan \phi = x \) and \( \cot \phi - \cot \theta = y \).

### Step-by-step Solution:

1. **Rewrite the given equations**:
We have:
\[ \tan \theta - \tan \phi = x \]
\[ \cot \phi - \cot \theta = y \]

2. **Express \(\tan\) in terms of \(\cot\)**:
Recall that \( \tan \theta = \frac{1}{\cot \theta} \) and \( \tan \phi = \frac{1}{\cot \phi} \). Therefore, we can rewrite the first equation:
\[ \frac{1}{\cot \theta} - \frac{1}{\cot \phi} = x \]

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

4. **Relate to the second equation**:
From the second equation, we have:
\[ \cot \phi - \cot \theta = -y \]
Substituting this into our equation gives:
\[ \frac{-y}{\cot \theta \cdot \cot \phi} = x \]

5. **Rearranging gives us**:
\[ \cot \theta \cdot \cot \phi = -\frac{y}{x} \]

6. **Use the cotangent subtraction formula**:
The formula for \( \cot(\theta - \phi) \) is:
\[ \cot(\theta - \phi) = \frac{\cot \theta \cdot \cot \phi + 1}{\cot \phi - \cot \theta} \]

7. **Substituting values**:
We already found \( \cot \theta \cdot \cot \phi = -\frac{y}{x} \) and \( \cot \phi - \cot \theta = -y \). Substituting these into the formula gives:
\[ \cot(\theta - \phi) = \frac{-\frac{y}{x} + 1}{-y} \]

8. **Simplifying**:
This can be simplified as:
\[ \cot(\theta - \phi) = \frac{1 - \frac{y}{x}}{y} = \frac{x - y}{xy} \]

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