1. cos (pi/2 - x) tan x = a. sin x tan x b. cos x c. tan x d. cosec x e. sec x 2. sec alpha tan2 alpha + sec alpha a. sec alpha b. sec2 alpha c. sec3 alpha d.
Matematika
michiireika
Pertanyaan
1. cos (pi/2 - x) tan x =
a. sin x tan x
b. cos x
c. tan x
d. cosec x
e. sec x
2. sec alpha tan2 alpha + sec alpha
a. sec alpha
b. sec2 alpha
c. sec3 alpha
d. cosec alpha
e. cosec2 alpha
3. (tan pi/4 + tan pi/6) / (1 - tan pi/4 tan pi/6)
a. akar 3
b. 2 - akar 3
c. 2 + akar 3
d. 2akar3 - 1
e. 2akar3 +1
a. sin x tan x
b. cos x
c. tan x
d. cosec x
e. sec x
2. sec alpha tan2 alpha + sec alpha
a. sec alpha
b. sec2 alpha
c. sec3 alpha
d. cosec alpha
e. cosec2 alpha
3. (tan pi/4 + tan pi/6) / (1 - tan pi/4 tan pi/6)
a. akar 3
b. 2 - akar 3
c. 2 + akar 3
d. 2akar3 - 1
e. 2akar3 +1
1 Jawaban
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1. Jawaban bapakdr
pi = 180
1) cos (180/2 - x) tan x
= cos (90-x) tan x
= sin x tan x
2) sec a tan²a + sec a
= sec a (sec² - 1) + sec a
= sec³a - sec a + sec a
= sec³a
3) (tan 180/4 + tan 180/6)/(1 - tan 180/4 tan 180/6)
= (tan 45 + tan 30)/(1 - tan 45 tan 30)
= (1 + ⅓√3) / (1 - 1(⅓√3))
= (3/3 + √3/3) / (3/3 - √3/3)
= (3+√3)/3 / (3-√3)/3
= (3+√3)/3 × 3/(3-√3)
= (3+√3)/(3-√3)
= (3+√3)/(3-√3) × (3+√3)/(3+√3)
= (9+6√3+3) / (9-3)
= (12+6√3) / 6
= 2+√3