Finding the value of $\tan B$

Question

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!

Answer 1

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...

Answer 2

$$\begin{align*} 1&=\sin A\cos B-\cos A\cos B\\[1ex] &=\cos A\cos B(\tan A-1)\\[1ex] &=\cos A\cos B\left(\frac1{\cot A}-1\right)\\[1ex] &=-\frac73\cos A\cos B \end{align*}$$

$$\implies\cos B=-\frac37\sec A\quad(*)$$

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

$$\tan B=\sqrt{\frac{\sin^2B}{\cos^2B}}=\frac{2\sqrt6}5$$