If tanA=2/3 and sinB=5/7 and angles A and B are in Quadrant 1 fine the value of tan(A+B)?

Question

I used the tan(a+b)=tanA+tanB/1-tanA times tanB formula. But how do I find tan B and what does the answer come out to?

Answer 1

tanA = 2 / 3 and sinB = 5 / 7 and angles A and B are first quadrant then

Recall : Trigonometry identities

cosB = √24 / 7

Therefore tanB = sinB / cosB = 5 / 7 / √24 / 7 = 5 / √24

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

Substitute tanA = 2 / 3 and tanB = 5 / √24 in the above formula

                   = (2 / 3 + 5 / √24 ) / [1 - (2 / 3)(5 / √24)]

                   = (2√24 + 15) / 3√24  / [1 - 10 / 3√24 ]   

                   = (2√24 + 15) / [3√24 - 10]

                   = (2√24 + 15)[3√24 + 10] / [3√24 + 10] [3√24 - 10]

                   = (144 +20√24 + 45√24 + 150) / (216 - 100)

                   = (294 + 65√24) / 116

                   = (147 + 65√6) / 58.