# 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.

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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.` **