Question

Помогите с решением, пожалуйста

Answer 1

1) cos(2α+810°)=cos(720°+90°+2α)=cos(360°*2+90°+2α)=
   =cos(90°+2α)=-sin2α

2) sin(2α+450°)=sin(360°+90°+2α)=sin(90°+2α)=cos2α

3) cos(2α-630°)=cos(-(630°-2α))=cos(630°-2α)=cos(720°-90°-2α)=
    =cos(360°*2-90°-2α)=cos(-90°-2α)=cos(-(90°+2α))=cos(90°+2α)=
    =-sin2α

4) sin(2α+990°)=sin(1080°-90°+2α)=sin(360°*3-90°+2α)=sin(-90°+2α)=
   =sin(-(90°-2α))=-sin(90°-2α)=-cos(-2α)=-cos2α

1-cos(2α+810°)+sin(2α+450°)=
1-cos(2α-630°)+sin(2α+990°)

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

Answer 2

[1+cos(2a+90)+sin(2a+90)]/[1-cos(2a+90)+sin(2a-90)]=
=(1+sin2a+cos2a)/(1+sin2a-cos2a)=
=(sin²a+cos²a+2sinacosa+cos²a-sin²a)/(sin²a+cos²a+2sinacosa-cos²a+sin²a)=
=(2cos²a+2sinacosa)/(2sin²a+2sinacosa)=
=2cosa(cosa+sina)/2sina(sina+cosa)=ctga