Question

a) sen 135°+ tg 300°
b) cotg 240°- sec 150°
c) cos 60°× cossec 210°
d) cos 225°÷ cotg 30°​

Answer 1

Explicação passo-a-passo:

a) Sen 135° = Sen 45° = √2/2

tag 300° = -tag(360-300)

tag 300° = -tag(60)= -√3

√2/2 - √3 = √2/2 -2√3/2= (√2 - 2√3)/2

b) cotg 240 = 1/tag(240) = 1/√3 = √3/3

sec 150 = 1/ Cos(150)= 1/(-√3/3)= -3/√3 = -√3

√3/3 -( -√3) = √3/3 + √3 = 4√3/3

c) Cos 60° = √3

cossec 210° = - cossec 30

- cossec 30 = - 1/ Cos 30 = -1/ (√3/2)

cossec 210 = -2√3/3

√3.(-2√3)/3 = -2 3/3= -2

d) Cos 225 = - √2/2

cotg 30 = √3

-√2/2 ÷ √3 = - √6/6