LetA be in Q1, cos = -5/13, let B be in q2 & tanB=-2, find tan(a+b) and determine in which Quadrant A+B terminates. Draw 3 diagrams.
The problem provides an information that is not correct: cos A can not be negative `-5/13` since A is in quadrant 1.
Assuming that `cos A = 5/13` and `tan B = -2,` you need to find the tangent of sum of angles A and B such that:
`tan(A+B) = (tan A + tanB)/(1 - tanA*tanB)`
Since you do not know the value of tan A, you may find it using the following formula:
`cos^2 A = 1/(1 + tan^2 A)`
`25/169 = 1/(1 + tan^2 A)`
`25/169 + 25/169*tan^2 A = 1`
`25/169*tan^2 A = 1 - 25/169`
`25/169*tan^2 A = 144/169`
`25 tan^2 A = 144 =gt tan^2 A = 144/25`
`tan A = +-sqrt(144/25)`
Since A is in quadrant 1, then you will keep only the positive value for tan A such that:
`tan A = 12/5`
`tan(A+B) = (12/5 - 2)/(1 - 24/5)`
`tan(A+B) = -2/19`
Since the tangent is negative, the angle A+B is in quadrant 2, since in the values of tangent are positive in quadrant 3.
Hence, evaluating `tan(A+B)` yields `tan(A+B) = -2/19.`