Using the fact that cot(x)=1/tan(x) and the one you wrote,
\(\displaystyle \L \cot(A-B) = \frac{1}{\tan(A-B)}
= \frac{1}{\frac{\tan(A) - \tan(B)}{1+\tan(A) \tan(B)}}
= \frac{1+\tan(A) \tan(B)}{\tan(A) - \tan(B)}\)
It is desirable to express this in terms of cot(A) and cot(B),
\(\displaystyle \L = \frac{1+\frac{1}{\cot(A)} \frac{1}{\cot(B)}}{\frac{1}{\cot(A)} - \frac{1}{\cot(B)}}=?\)
You get it after simplifying.
