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)=(\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((20)/(29)\right)^2}=-(21)/(29)`

So we have


`\tan B=(\sin B)/(\cos B)=-(20)/(21)`

So,


`\tan(A+B)=(\frac13-(20)/(21))/(1+\frac13\cdot(20)/(21))=-(39)/(83)`