in Q3, tan is positive, so tanA = √48
in Q2, tan is negative, so tanB = -1/√3
Now apply the addition formula
tan(A+B) = (tanA+tanB)/(1-tanA tanB)
Evaluate the expression under the given conditions.
tan(A+B); cosA=-1/7, A in quadrant 3, sinB=1/2, B in quadrant 2.
1 answer