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

$\begin{array}{1 1}(a)\;\large\frac{1}{y}-\frac{1}{x}&(b)\;\large\frac{1}{x}-\frac{1}{y}\\(c)\;\large\frac{1}{x}+\frac{1}{y}&(d)\;None\;of\;these\end{array}$

We have
$\tan A-\tan B=x$
$\cot B-\cot A=y$
Now, $\cot B-\cot A=y$
$\large\frac{\tan A-\tan B}{\tan A\tan B}$$=y \large\frac{x}{\tan A\tan B}$$=y$
$\tan A\tan B=\large\frac{x}{y}$
Now,$\cot(A-B)=\large\frac{1}{\tan(A-B)}$
$\Rightarrow \large\frac{1+\tan A\tan B}{\tan A-\tan B}$
$\Rightarrow \large\frac{1+x/y}{x}$
$\Rightarrow \large\frac{x+y}{xy}$
$\Rightarrow \large\frac{1}{x}+\frac{1}{y}$
Hence (c) is the correct answer.