tanA917 secB83 pi

tan(A)=9/17 sec(B)=8/3 pi<A<3pi/2<B<2pi

sin(A+B)

Solution

tan A= 9/17    A is in third quadrant

Here opposite= 9       and   adjacent   =17

Therefore hypotenuse= sqrt(17²+9²)=sqrt 370

sin A= - 9/sqrt370

cos A= -17/sqrt370

sec B=8/3

cos B=1/sec B= 3/8

here adjacent=3   and hypotenuse=8

opposite=sqrt(8²-3²)= sqrt55

sin B= - sqrt55/8

sin(A+B)= sinA cos B + cos A sinB

                = (-9/sqrt370)(3/8) + (-17/sqrt370)(sqrt55/8)= -27/ 8sqrt370 - 17sqrt55/8sqrt370

                                                                                          = (-27-17sqrt55)/8sqrt370