Question

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

Answer 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-√3
arctg(2-√3)=π/12
tgπ/12=2-√3
(1-cosπ/6)/sinπ/6=(1-√3/2)/(1/2)=
(2-√3):1/2=2-√3