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Question:

Let α and β be the roots of the quadratic equation x²sinθ - x(sinθcosθ + 1) + cosθ = 0 (0 < θ < 45°), and α < β. Then ∞∑n=0(αⁿ + (-1)ⁿβⁿ) is equal to:

1/(1 - cosθ) + 1/(1 + sinθ)

1/(1 + cosθ) + 1/(1 - sinθ)

1/(1 - cosθ) + 1/(1 + sinθ)

1/(1 + cosθ) + 1/(1 - sinθ)

Solution:

D = (1 + sinθcosθ)² - 4sinθcosθ = (1 - sinθcosθ)² ≥ 0.
Roots are β = cosθ and α = sinθ ⇒ ∞∑n=0(αⁿ + (-1)ⁿβⁿ) = ∞∑n=0(cosθ)ⁿ + ∞∑n=0(-sinθ)ⁿ = 1/(1 - cosθ) + 1/(1 + sinθ)