Question 1042126
Tan(A+B) = (TanA+TanB)/(1-TanATanB)
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We know that Tan = 1/cot

If (a+b) = 90 then tan(a+b)= tan 90
We know that tan 90 = 0 then tan(a+b)= 0

We know that  Tan(A+B) = (TanA+TanB)/(1-TanATanB)= 0 for 
(TanA+TanB)/(1-TanATanB)to be equal to zero the numerator has to be equal to zero which means  TanA+TanB = 0. We already know what the tan of a is (Tan a = 1/CotA) This means that tan a = 1/(4/3) = 3/4 (Please note that the problem has already given us the value of Cot A which is 4/3

Therefore Tan A + Tan B = 0 replacing Tan A with 3/4 we get
Tan B + 3/4 = 0. Therefore Tan B = -3/4