Question

Помогите с 1 и 4 пожалуйста

Answer 1

1)
cos(-π/3)=cos(π/3)=1/2
sin(π/6)=1/2
tg(π/4)=1
sinπ=0
ctg(π/6)=√3

(4·(1/2)+2·(1/2)-5):(0+2√3)=-3/2√3=-√3/2

4)
tg(π+β)=tgβ
ctg((π/2)-β)=tgβ
ctg((3π/2)-β)=tgβ
tg(π-β)=-tgβ

tg(π+β)·ctg((π/2)-β)-ctg((3π/2)-β)·tg(π-β)=tgβ·tgβ-tgβ·(-tgβ)=2tg²β