[tex]\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{\tan A+\frac43}{1-\frac43\tan A}[/tex]
[tex]\cos^2A=1-\sin^2A\implies\cos A=\pm\sqrt{1-\sin^2A}[/tex]
where the sign of the root is chosen depending on the value of [tex]A[/tex]. Since we know [tex]90^\circ<A<180^\circ[/tex], it follows that [tex]-1<\cos A<0[/tex], i.e. [tex]\cos A<0[/tex], which means
[tex]\cos A=-\sqrt{1-\sin^2A}=-\dfrac5{13}[/tex]
[tex]\implies\tan A=\dfrac{\sin A}{\cos A}=\dfrac{\frac{12}{13}}{-\frac5{13}}=-\dfrac{12}5[/tex]
[tex]\implies\tan(A-B)=\dfrac{-\frac{12}5+\frac43}{1+\frac43\times\frac{12}5}=-\dfrac{16}{63}[/tex]
