Question

Помогите решить плиз!!!

Answer 1

12)cosα /(1+sinα)+tgα=cosα /(1+sinα)+sinα/cosα =
=(cos²α+sinα+sin²α)/[(1+sinα)·cosα] =(1+sinα)/[(1+sinα)·cosα] =1/cosα

3) ctgα=-2,  π/2<α<π

α=arcctg(-2)=π-arcctg(2)

17)sinα=-2/√29,  3π/2<α<2π
α=arcsin(-2/√29)=-arcsin(2/√29)

2)1-(sinαcosα)/ctgα=1-(sinαcosα)/(cosα/sinα)=1-sin²α cosα/cosα=
=1-sin²α=cos²α