$\tan x=\tan \left( a-B \right)=\frac{\tan a-\tan B}{1+\tan a\tan B}=\frac{\frac{8}{4}-\frac{2}{4}}{1+\left( \frac{8}{4} \right)\left( \frac{2}{4} \right)}=\frac{\frac{6}{4}}{2}=\frac{3}{4}$