There are some relations between the trigonometric functions. Tan(A-B) is only one of them and it is expressed by the following:
[tex]\begin{gathered} \tan (A-B)=\frac{(\tan A-\tan B)}{(1+(\tan A)\cdot(\tan B))} \\ \tan (A-B)=\frac{(\frac{4}{3}-\frac{12}{5})}{(1+(\frac{4}{3})\cdot(\frac{12}{5}))} \end{gathered}[/tex]So, to get the answer we just solve the arithmetic operations as following:
[tex]\begin{gathered} \tan (A-B)=\frac{(\frac{20-36}{15})}{(1+(\frac{48}{15}))} \\ \tan (A-B)=\frac{(\frac{-16}{15})}{(\frac{5}{5}+(\frac{16}{5}))} \\ \tan (A-B)=\frac{\frac{-16}{15}}{\frac{21}{5}} \\ \tan (A-B)=\frac{-16\cdot5}{15\cdot21} \\ \tan (A-B)=\frac{-80}{315} \\ \tan (A-B)=-\frac{16}{63} \end{gathered}[/tex]