Finding the value of $\tan B$ 
In the triangle $\triangle ABC$
$$\cot A=\dfrac{-3}{4}$$
$$\sin A \cos B - \cos A \cos B = 1 $$

*

*Find the value of $\tan B$

This $\sin A \cos B - \cos A \cos B = 1 $ reminds me of sum and difference formulas in a particular way that made me draw a triangle. However, I couldn't see any way to proceed from there. Your helps will be appreciated.
Regards!
 A: Hint: we have  $$\cot A = {\cos A\over \sin A}$$ and $$\sin ^2A +\cos ^2 A =1$$
so $\cos A =-3t$ and $\sin A = 4t$ for some real $t$ and thus $25t^2=1$ so $t=1/5$ (because $\sin A$ must be positive).  
Now $\displaystyle \cos B = -{1\over 7t} =-{5\over 7}$ so $\sin B = ...$ and so...
A: $$\begin{align*}
1&=\sin A\cos B-\cos A\cos B\\[1ex]
&=\cos A\cos B(\tan A-1)\\[1ex]
&=\cos A\cos B\left(\frac1{\cot A}-1\right)\\[1ex]
&=-\frac73\cos A\cos B
\end{align*}$$
$$\implies\cos B=-\frac37\sec A\quad(*)$$
Then by the Pythagorean identities, we have
$$\begin{align*}
\sin^2B&=1-\cos^2B\\[1ex]
&=1-\left(-\frac37\sec A\right)^2\\[1ex]
&=1-\frac9{49}\sec^2A\\[1ex]
&=1-\frac9{49}(\tan^2A+1)\\[1ex]
&=\frac{40-9\tan^2A}{49}\\[1ex]
&=\frac{40\cot^2A-9}{49\cot^2A}\\[1ex]
&=\frac{24}{49}
\end{align*}$$
$$\implies\cos^2B=1-\frac{24}{49}=\frac{25}{49}$$
Since $A,B$ are angles in a triangle, we have $0<A<\pi$ and $0<B<\pi$. Then we know that $\sin B>0$. Similarly, we know that $\sin A>0$, while $\cot A<0\implies\cos A<0\implies\sec A<0$. This and $(*)$ mean that $\cos B>0$, and taken together we know that $\tan B>0$.
So, we end up with
$$\tan B=\sqrt{\frac{\sin^2B}{\cos^2B}}=\frac{2\sqrt6}5$$
