高一三角函数题2tan[arcsin(3/4)]=?
tan[arin(3/4)]=?
sin[arcsin(3/4)]=3/4, 则arcsin(3/4)∈(0,π/2) ==>cos[arcsin(3/4)]=√7/4 所以 1+{tan[arcsin(3/4)]}^2=1/{ cos[arcsin(3/4)]}^2 tan[arcsin(3/4)]=(3√7)/7