[tex]\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}[/tex]
Since [tex]B[/tex] is quadrant 2, we expect [tex]\cos B<0[/tex]. Then
[tex]\cos^2B+\sin^2B=1\implies\cos B=-\sqrt{1-\left(\dfrac{20}{29}\right)^2}=-\dfrac{21}{29}[/tex]
So we have
[tex]\tan B=\dfrac{\sin B}{\cos B}=-\dfrac{20}{21}[/tex]
So,
[tex]\tan(A+B)=\dfrac{\frac13-\frac{20}{21}}{1+\frac13\cdot\frac{20}{21}}=-\dfrac{39}{83}[/tex]
