Given \[ \tan A - \tan B = x \quad \text{...(i)} \] and \[ \cot B - \cot A = y \quad \text{...(ii)} \] From Eq(ii), \[ \cot B - \cot A = y \] we get \[ \frac{1}{\tan B} - \frac{1}{\tan A} = y \] \[ \frac{\tan A - \tan B}{\tan B \tan A} = y \] \[ \frac{x}{\tan B \tan A} = y \] Thus, \[ \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \] \[ = \frac{\frac{y}{x} + 1}{y} \quad \text{(using Eq(ii) and Eq(iii))} \] \[ = \frac{1}{x} + \frac{1}{y} \]