arctg(x)=p/24 если х=√6-√3+√2-2
и
arctg(x)=p/12 если х=(√3-1)/(√3+1)

1)х=√6-√3/(√2-2)=√3(√2-1)-√2(√2-1)=(√2-1)(√3-√2)tgπ/24=(√2-1)(√3-√2)2)x=(√3-1)/(√3+1)=(√3-1)²/(3-1)=(3-2√3+1)/2=(4-2√3)/2=2-√3arctg(2-√3)=π/12tgπ/12=2-√3(1-cosπ/6)/sinπ/6=(1-√3/2)/(1/2)=(2-√3):1/2=2-√3