Question

tg(arcsin3/5-arccos1/4) помогите!! 

Answer 1

Обозначим
arcsin 3/5=α, тогда sin α=3/5,  0≤α≤π/2
найдем cos α=√1 - (3/5)²=√1- 9/25=√16/25=4/5     и tg α=sin α/cos α=3/4

arccos 1/4=β, cos β=1/4  b 0≤β≤π/2
sin β=√1-(1/4)²=√15/4  и tg β=√15

tg(arcsin 3/5-arccos 1/4)=tg(α-β)=(tg α - tgβ)/1+tgα·tgβ=(3/4-√15)/1+3√15/4 =
=(3-4√15)/(4+3√15)