If $\tan A-\tan B=x$  and $\cot B-\cot A=y$  then $\cot (A-B)$  is

$\begin{array}{l}\tan A-\tan B=x\\ \Rightarrow \frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}=x\\ \Rightarrow \sin A.\cos B-\sin B.\cos A=x\cos A.\cos B\\ \Rightarrow \sin (A-B)=x\cos A.\cos B\cdots \left(1\right)\\ \cot B-\cot A=y\end{array}$

$\begin{array}{l}\frac{\cos B}{\sin B}-\frac{\cos A}{\sin A}=y\\ \sin A\cos B-\cos A\sin B=y\sin A\sin B\\ \sin (A-B)=y\sin A\sin B\cdots \left(2\right)\end{array}$

From $\left(1\right)\&\left(2\right)$

$x\cos A.\cos B=y\sin A.\sin B$

$\cot A.\cot B=\frac{y}{x}$

$\cot (A-B)=\frac{\cot A.\cot B+1}{\cot B-\cot A}$

$=\frac{\frac{y}{x}+1}{y}=\frac{y+x}{xy}$

$=\frac{1}{x}+\frac{1}{y}$