$If\tan A-\tan B=xand\cot B-\cot A=y, then\cot (A-B)\ is\ equal\ to$
$\left(A\right).\ \frac{1}{x}-\frac{1}{y}\ \ \left(B\right).\ \frac{1}{x}+\frac{1}{y}\ \left(C\right).\ \frac{1}{y}-\frac{1}{x}\ \left(D\right).\ \frac{1}{x+y}$
from the equation,
$\cot B-\cot A=y$
$\frac{1}{\tan B}-\frac{1}{\tan A}\ =\ y$
$\frac{\tan A-\tan B}{\tan A\ \tan B}\ =\ y$
$\frac{x}{\tan A\ \tan B}\ =\ y$
$\tan A\tan B\ =\ \frac{x}{y}$
$\therefore \cot A\ \cot B\ =\ \frac{y}{x}$
Now use the below formula, to get the correct solution.
$\cot (A−B)=\frac{\left(\cot A\cot B+1\right)}{\left(\cot B−\cot A\right)}$
$\cot (A−B)=\frac{\left(\frac{y}{x}+1\right)}{\left(y\right)}$
$=\ \frac{\left(\frac{y+x}{x}\right)}{y}$
$=\frac{y+x}{xy}$
$=\frac{y}{xy}+\frac{x}{xy}$
$\cot \ \left(A-B\right)=\frac{1}{x}+\frac{1}{y}$