Question

помогите пожалуйста 2 и 6

Answer 1

2) 1/(sinα - 1) - 1/(sinα + 1) = (sinα + 1- sinα + 1)/(sin²α - 1) = -2/cos²α
6) cos²α - (ctg²α + 1)· sin²α = cos²α - (cos²α/sin²α + 1)· sin²α =
= cos²α - (cos²α + sin²α) = -sin²α

Answer 2

2.
1/(sinα-1)-1/(sinα+1)=(sinα+1-(sinα-1)/((sinα+1)(sinα-1))=
=(sinα+1-sinα+1)/(sin²α-1)=2/(-cos²α)=-2*arccos²(α).
6.
cos²α-(ctg²α+1)*sin²α=cos²α-(cos²α/sin²α+1)*sin²α=cos²α-(cos²α+sin²α)=
cos²α-1=-sin²α.