0 < a < (π/2) → di kuadran 1 (3π/2) < b < 2π → di kuadran IV
sin a = 3/5 = y/r
x = √(r² - y²) x = √(5² - 3²) x = √(25 - 9) x = √16 x = 4 tan a = y/x = 3/4
cos b = 7/25 = x/r
y = √(r² - x²) y = √(25² - 7²) y = √625 - 49) y = 24 tan b = y/x = -24/7
cot (a - b) = 1 / (tan (a - b))
tan (a - b) = (tan a - tan b)/(1 + tan a . tan b) tan (a - b) = (3/4 - (-24/7)) / (1 + 3/4 . (-24/7)) tan (a - b) = (3 . 7 + 4 . 24)/(4 . 7) : (1 - 18/7) tan (a - b) = (21 + 96)/(28) : (-11/7) tan (a - b) = 117/28 x (-7/11) tan (a - b) = -117/44
Verified answer
Bab TrigonometriMatematika SMA Kelas X
α = a
β = b
0 < a < (π/2) → di kuadran 1
(3π/2) < b < 2π → di kuadran IV
sin a = 3/5 = y/r
x = √(r² - y²)
x = √(5² - 3²)
x = √(25 - 9)
x = √16
x = 4
tan a = y/x = 3/4
cos b = 7/25 = x/r
y = √(r² - x²)
y = √(25² - 7²)
y = √625 - 49)
y = 24
tan b = y/x = -24/7
cot (a - b) = 1 / (tan (a - b))
tan (a - b) = (tan a - tan b)/(1 + tan a . tan b)
tan (a - b) = (3/4 - (-24/7)) / (1 + 3/4 . (-24/7))
tan (a - b) = (3 . 7 + 4 . 24)/(4 . 7) : (1 - 18/7)
tan (a - b) = (21 + 96)/(28) : (-11/7)
tan (a - b) = 117/28 x (-7/11)
tan (a - b) = -117/44
cot (a - b) = 1 : -117/44 = -44/117