Question

Помогите, пожалуйста, срочно.

Answer 1

(1-sinα)(1+sinα) = 1 - sin²α = cos²α

(sinα - cos α)² + 2sinαcosα=sin²α+cos²α-2sinαcosα+2sinαcosα=1

(sin²α)/(1+cosα) + cosα = (sin²α+cos²α+cosα)/(1+cosα)=(1+cosα)/(1+cosα)=1

cos(π/2+α) + sin(π+α) = -sinα-sinα = -2sinα =-2sin(π/3) = -2 * √3/2 = -√3

sin(π+α)/sin(π/2+α) = -sinα / cosα = -tgα = -tg(π/4) = -1

tg(π-α)+tg(-α)=-tgα-tgα = -2tgα = -2 * tg(π/4) = -2 * 1= -2

sin²α/(1-cosα) - cosα = (1-cos²α)/(1-cosα) - cosα = (1-cosα)(1+cosα)/(1-cosα) - cosα = 1 + cosα - cosα = 1

(1+tgα)/(1+ctgα) = (tgα+tg²α)/(tgα+1) = tgα(tgα+1)/(tgα+1) = tgα