Ответ:

Объяснение:

sin(90°+α)=cosα     cos(90°+α)=-sinα

=> sin130°= sin (90°+40°)=cos 40°        sin 110°=cos20°

cos130°= cos (90°+40°)=-sin 40°        cos 110°= -sin 20°

=> [tex]\frac{sin130+sin110}{cos130+cos110}=\frac{cos40+cos20}{-sin40-sin20}= -\frac{cos40+cos20}{sin40+sin20}[/tex]

cosα+cosβ= 2cos((α+β)/2)cos((α-β)/2)     sinα+sinβ=2sin((α+β)/2)cos((α-β)/2)

[tex]= > -\frac{cos 40+cos20}{sin40+sin20}= -\frac{2cos((40+20)/2)*cos((40-20)/2)}{2sin((40+20)/2)*cos((40-20)/2)} =-\frac{cos30*cos10}{sin30*cos10}= -\frac{cos30}{sin30}= -\frac{\sqrt{3}}{2}: \frac{1}{2} =-\sqrt{3}[/tex]