Question

Решите плиз срочно нужно

Answer 1

1)a)√3(cos75°-cos15°)=√3(-2sin45°•sin30°)
=-2√3•√2/2•1/2= -√6/2
b)1-2sin^2(15°)=1-(1-cos30°)=cos30°=√3/2
c)-√6/2:√3/2=-√2
2)sin35°+sin25°=cos5°
sin35°+sin25°=2•sin(35°+25°)/2•
cos(35°-25°)/2=2•sin30°•cos5°=
2•1/2•cos5°=cos5°
3)1+tga•ctgb=cos(a-b)/(cosa•cosb)
cos(a-b)/(cosa•cosb)=
(cosacosb+sinasinb)/(cosacosb)=
1+(sinasinb)/(cosacosb)=1+tgatgb