please bantu jawab dong

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12. nilai maksimum
f(x) = 12/ { 8 -√15 cos x +sin x)
f(x)= 12 / {8 + (-√15 cos x + sin x)}
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-√15 cos x + sin x = k. cos (x -α)
k² =(-√15)² +(1)²
k² = 16
|k|  = 4

f(x)= 12/ {8 + 4 cos (x - α)}
dan -1 ≤ cos (x-α) ≤ 1
untuk cos (x - α)= 1 --> f(x) = 12/ { 8+4(1)} = 12/12 =1
untuk cos (x - α) = 1--> f(x)=  12/{8+ 4(-1)} = 12/4 = 3
nilai maksimum f(x)= 3

13. perkalian matriks A(2x2). B(2x2)
b = baris
k = kolom

b1k1 = ³log 5. ⁵log 9 + secθ. cosθ=³log 9 + 1/cos θ. cos θ=  2 + 1 = 3
b1k2 = ³log 5.⁵log 3 + sec θ(tan θ sec θ)= ³log3 + tan θ sec² θ= 1+ tan θsec²θ
b2k1 = ⁹log5. ⁵log 9 -cos θ (cos θ) = ⁹log 9 - cos²θ = 1-cos² θ= sin²θ
b2k2 = ⁹log5.⁵log 3 - cos θ(tanθ secθ) = ⁹log 3 - tanθ = 1/2 - tan θ
=
|3..........1+tan θ sec² θ|
|sin² θ..........1/2 - tan θ|