If tanα and tanβ are roots of equation x² – 6x + 8 = 0, then what is cos(2α + 2β)?

Sum of roots: tanα + tanβ = 6. Product: tanα·tanβ = 8. tan(α+β) = (tanα+tanβ)/(1–tanα tanβ) = 6/(1–8) = –6/7. Then tan(2α+2β) = 2tan(α+β)/(1–tan²(α+β)) = 2(–6/7)/(1 – 36/49) = (–12/7)/(13/49) = (–12/7)(49/13) = –84/13. cos(2α+2β) = 1/√(1+tan²) with sign. cos²(2α+2β) = 1/(1+7056/169) = 169/7225. cos(2α+2β) = ±13/85. To determine sign: with tan(2α+2β) = –84/13 < 0, the angle is in Q2 or Q4. Standard convention gives cos = –13/85 or +13/85. The answer is 13/85, but with the sign analysis, it's –13/85; given options, (c) 17/85 doesn't match. Likely answer is (b) 13/85.