Question

Помогите решить букву б

Answer 1

б)
cos²α-(cos(π-α)*sin(π/2-α))/(ctg(π+α)*tg(3π/2-α))=           ОДЗ: α≠πn/2, n∈Z.
=cos²α-(-cosα*cosα)/(ctgα*ctgα)=cos²α-(-cos²α/(cos²α/sin²α))=cos²α+sin²α=1.