Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$ The question is:

Prove that:
  $$ \cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B} $$

I have tried expanding it as $\dfrac{\cos(A+B)}{\sin(A+B)}$ and $\dfrac{1}{\tan(A+B)}$.

It is probably easier to start from the RHS:
$$\frac{\cot A\cot B-1}{\cot A+\cot B}= \frac{\frac{\cos A}{\sin A}\frac{\cos B}{\sin B} -1}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B} }$$
Now, all the following computations are natural, and pretty authomatic:
$$=\frac{\frac{\cos A \cos B}{\sin A \sin B}-\frac{\sin A\sin B }{\sin A\sin B} }{\frac{\cos A \sin B+ \sin A \cos B}{\sin A \sin B}}=\frac{\cos A \cos B-\sin A\sin B }{\cos A \sin B+ \sin A \cos B}=\frac{\cos(A+B)}{\sin(A+B)}$$
