Solving for Sin/cos/tan(A+B)

Question

Hello all i have been puzzling on this for several hours i got Cos(A+B) correct but cannot find sin or tan. 

 

Here is the problem Let 

sin A = 

12

13

 with A in QII and 

sin B = − 

4

5

 with B in QIII. Find sin(A + B), cos(A + B), and tan(A + B)

 

My answers so far are sin(A+B) = -16/65 

Cos(A+B) = 63/65

tan(A+B) = -16/63

Thanks!

Answer 1

A lies in 2nd quadrant ; SinA is +ve, CosA is - ve, tanA is - ve.

  B lies in 3rd quadrant ; SinB is - ve, CosB is -ve, tanB is + ve

sinA = 12/13 ---> cosA = - 5/13, tanA = -12/5.

sinB = -4/5 ---> cos B = - 3/5, tanB = 4/3

1.

sin(A+B) = sinA*cosB + cosA*sinB

                   = (12/13 * - 3/5 ) + (-5/13 * -4/5)

                   = -36/65   + 20/65

                   = -16/65

2.

cos(A+B) = cosA*CosB - sinA*sinB

                   = (-5/13 * - 3/5 ) - (12/13 * -4/5)

                   = 15/65  +  48/65

                   = 63/65

Answer 2

Given that sin A =

From triangle,

cos A =

Given that sin B =

cos B =

Answer 3

3.

tan(A+B) = (tanA + tanB) / ( 1 - tanA*tanB)

             = (-12/5 + 4/3 ) - (1 - (12/5) * (4/3))

             = (-16/15) / (1 + 48/15)

             = (-16 /15) / (63 / 15)

             = -16/63