Question

Помогитее решить пожалуйста

1) tg(180°-α)/ctg(90°-α)
2) cos²(90°-α)-1/cos(180°-α)
3)sin(π-α)/tg(π+α)
4)tg(π-α)/ctg(π/2-α)



1)tg(π-α)
2)cos(360°-α)
3)sin(90°-α)-cos(180°-α)+tg(180°+α)-ctg(270°-α)
4)ctg(360°-α)
5)ctg(π+α)
6)sin(360°+α)
7)tg(360°+α)

Answer 1

1)tg(180°-α)/ctg(90°-α)=(-tgα)/tgα=-1;
2)(cos²(90°-α)-1)/cos(180°-α)=(sin²α-1)/(-cosα)=-cos²α/-cosα=cosα;
3)sin(π-α)/tg(π+α)=sinα/tgα=sinα/(sinα/cosα)=cosα;
4)tg(π-α)/ctg(π/2 -α)=-tgα/tgα=-1;
1)tg(π-α)=-tgα;
2)cos(360°-α)=cosα;
3)sin(90°-α)-cos(180°-α)+tg(180°+α)-ctg(270°-α)=
=cosα-(-cosα)+tgα-tgα=2cosα;
4)ctg(360°-α)=-ctgα;
5)ctg(π+α)=ctgα;
6)sin(360°+α)=sinα;
7)tg(360°+α)=tgα.