Question

помогите а и б) пожалуйста.

Answer 1

1) (sinα·cosα - tgα)/(1-(sinα+cosα)²)=(1/2)tg²α

((sinα·cosα - tgα)/(1-(sinα+cosα)²)=
=(sinα·cosα - sinα/cosα)/(1-(sin²α+2sinα·cosα+cos²α))=
[(sinα·(cos²α - 1))/cosα]/(1-1-2sinα·cosα)= (- sin³α)/(-2sinα·cosα)=(1/2)tg²α
ч.т.д.

2) [(sinα+cosα)² -1]/( ctgα-sinα·cosα)=2tg²α

[(sin²α+2sinαcosα+cos²α)-1]/( cosα/sinα-sinα·cosα)=
(1+2sinαcosα-1)/[(cosα-sin²α ·cosα)/sinα]=2sinαcosα/[(cosα(1-sin²α)/sinα]=2sin²αcosα/[(cosα(cos²α)]=2sin²α/(cos²α)=2tg²α

чтд