Помогите решить букву б
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.