(i) If `tanA-tan B=x` and `cot B- cot A=y`. Prove that `cot(A-B)=1/x + 1/y`. (ii) If `2 cos A = x+1/x, 2 cos B = y + 1/y`, then show that `2 cos (A-B)= x/y+y/x`.

Answer

Let's solve the problem step by step.

### Part (i)
We need to prove that \( \cot(A - B) = \frac{1}{x} + \frac{1}{y} \) given \( x = \tan A - \tan B \) and \( y = \cot B - \cot A \).

1. **Start with the cotangent difference formula:**
\[ \cot(A - B) = \frac{1}{\tan(A - B)} \]

2. **Use the tangent difference formula:**
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

3. **Substituting for \( x \):**
\[ \tan(A - B) = \frac{x}{1 + \tan A \tan B} \]

4. **Thus, we have:**
\[ \cot(A - B) = \frac{1 + \tan A \tan B}{x} \]

5. **Now, express \( y \) in terms of \( \tan A \) and \( \tan B \):**
\[ y = \cot B - \cot A = \frac{1}{\tan B} - \frac{1}{\tan A} = \frac{\tan A - \tan B}{\tan A \tan B} = \frac{x}{\tan A \tan B} \]

6. **From this, we can express \( \tan A \tan B \):**
\[ \tan A \tan B = \frac{x}{y} \]

7. **Substituting this back into the cotangent expression:**
\[ \cot(A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{y + x}{xy} \]

8. **Finally, we can rewrite this as:**
\[ \cot(A - B) = \frac{1}{x} + \frac{1}{y} \]

### Conclusion for Part (i):
Thus, we have proved that:
\[ \cot(A - B) = \frac{1}{x} + \frac{1}{y} \]