Trigonometry & Inverse Trigonometry Question - None

Question

If $\tan\alpha$ and $\tan\beta$ are the roots of the equation $x^2 - 3x - 2 = 0$ where $\alpha, \beta \in \left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$ and $\beta > \alpha$, then:

Accepted Answer

Step 1: Apply Vieta's formulas

From x² - 3x - 2 = 0, we have:

tan α + tan β = 3

tan α · tan β = -2

Step 2: Find tan(α+β)

Using tan(α+β) = (tan α + tan β)/(1 - tan α tan β)

tan(α+β) = 3/(1-(-2)) = 3/3 = 1

Step 3: Find α+β

Since 1 - tan α tan β = 3 > 0, we have α+β ∈ (-π/2, π/2)

tan(α+β) = 1 and α+β ∈ (-π/2, π/2) ⟹ α+β = π/4

Step 4: Find individual angles

The roots of x² - 3x - 2 = 0 are x = (3±√17)/2

Since tan α · tan β = -2 < 0, the roots have opposite signs

Since β > α and both ∈ (-π/2, π/2): α < 0 < β

tan α = (3-√17)/2 < 0 (negative root) ⟹ α = arctan((3-√17)/2) ∈ (-π/2, 0)

tan β = (3+√17)/2 > 0 (positive root) ⟹ β = arctan((3+√17)/2) ∈ (0, π/2)

∴ Answers: B (α+β = π/4) and C (α and β have opposite signs)

Question:

Step-by-Step Solution

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