Question

Find (tan A+B) if A and B are in quadrant 2 with tan A=(1/3) and sin B=(20/29)

Answer 1

$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

Since $B$ is quadrant 2, we expect $\cos B<0$. Then

$\cos^2B+\sin^2B=1\implies\cos B=-\sqrt{1-\left(\dfrac{20}{29}\right)^2}=-\dfrac{21}{29}$

So we have

$\tan B=\dfrac{\sin B}{\cos B}=-\dfrac{20}{21}$

So,

$\tan(A+B)=\dfrac{\frac13-\frac{20}{21}}{1+\frac13\cdot\frac{20}{21}}=-\dfrac{39}{83}$