If secα=5/3 and tanβ=15/8,?

tan(α - β)
= sin(α - β)/cos(α - β)
= (sinαcosβ - sinβcosα)/(cosαcosβ + sinαsinβ)
= (tanα - tanβ)/(1 + tanαtanβ)

If secα = 5/3,then
cosα = 1/secα = 1/(5/3) = 3/5.

If cosα = 3/5 and α is in quadrant IV, then
tanα = [-√(5^2 - 3^2)]/3 = -4/3

If tanα = -4/3 and tanβ = 15/8, then
tan(α - β)
= (tanα - tanβ)/(1 + tanαtanβ)
= (-4/3 - 15/8)/[1 + (-4/3)(15/8)]
= (-32/24 - 45/24)/(1 - 15/6)
= (-77/24)(-6/9)
= (-77/4)(-1/9)
= 77/36.