If secB=x,tanB=y(cosecA-cotA)=4/3,sinA+cosA=?

Answer 1

cscA-cotA = (1-cosA)/sinA = tan(A/2) = 4/3
so, tanA = 2tanA/(1 - tan^2A) = 4/3
3tanA = 2 - 2tan^2A
2tan^2A + 3tanA - 2 = 0
tanA = (-3±5)/4
tanA = -2 or 1/2

Now, knowing tanA, you can find sinA and cosA

Not sure what B has to do with anything there...