Find tan(A-B) if tanA and tanB are roots of x^2 - (1+k)x +k = 0, 0<B<A<90?

0 < B < A < 90°
故 0 < tanB < tanA < +∞.
k = tanA．tanB > 0

解方程式:
    0 = x^2 - (1+k)x + k = (x-1)(x-k)
或直接套公式:
tanA = {(1+k)+√[(1+k)^2-4k]}/2 
        = {(1+k)+|1-k|}/2 = max{1,k}
tanB = {(1+k)-|1-k|}/2 = min{1,k}

tan(A-B) = (tanA -tanB)/(1+tanA tanB)
    = (max{1,k}-min{1,k})/(1+k) 
    = |1-k|/(1+k)