Jika cos a = 1/3 untuk 3Π/2 < a < 2Π dan sin b = √2/3 untuk Π/2 < b < Π, maka sin(a+b)/(tg a+ tg b) sama dengan ....

## Pembahasan

Jawaban: A (-√7/9)

Menentukan nilai trigonometri:
- cos a = 1/3, 3π/2 < a < 2π (kuadran IV) → sin a = -√(1-1/9) = -√(8/9) = -2√2/3
- sin b = √2/3, π/2 < b < π (kuadran II) → cos b = -√(1-2/9) = -√(7/9) = -√7/3

tan a = sin a/cos a = (-2√2/3)/(1/3) = -2√2
tan b = sin b/cos b = (√2/3)/(-√7/3) = -√2/√7

sin(a+b):
= sin a·cos b + cos a·sin b
= (-2√2/3)(-√7/3) + (1/3)(√2/3)
= 2√14/9 + √2/9
= (2√14 + √2)/9 = √2(2√7+1)/9

tan a + tan b:
= -2√2 + (-√2/√7) = -√2(2 + 1/√7) = -√2(2√7+1)/√7

sin(a+b)/(tan a + tan b):
= [√2(2√7+1)/9] / [-√2(2√7+1)/√7]
= (1/9) × (-√7)
= -√7/9 = -(1/9)√7 ✓

Jawaban: A (-(1/9)√7) ✓