Question

arccos(cos230)+arcsin(sin(-400)+arctg(tg110)

Answer 1

arccos(cos230°) +arcsin(sin(-400°))+arctq(tq110°) =
arccos(-cos50°) +arcsin(-sin(360°+40°)) +arctq(tq180°-70°) =
=π - arccos(cos50°) - arcsin(sin40°) - arctq(tq70°)  =
π - 50π/180 - 40π/180 - 70π/180 =π/9.