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

For equation x² - 3x - 2 = 0 with roots tan α and tan β:

• tan α + tan β = 3

• tan α · tan β = -2

Step 2: Use tangent addition formula

tan(α + β) = (tan α + tan β)/(1 - tan α · tan β) = 3/(1 - (-2)) = 3/3 = 1

Step 3: Determine α + β

Since α, β ∈ (-π/2, π/2) and β > α, we have α + β ∈ (-π, π)

From tan(α + β) = 1 and α + β ∈ (-π, π), we get: α + β = π/4

Step 4: Find individual roots

From x² - 3x - 2 = 0: x = (3 ± √17)/2

Since β > α and both are in (-π/2, π/2):

• tan β = (3 + √17)/2 ≈ 2.56 → β = arctan((3 + √17)/2)

• tan α = (3 - √17)/2 ≈ -0.56 → α = arctan((3 - √17)/2)

Step 5: Verify conditions

✓ α + β = π/4

✓ β > α (since tan β > tan α and both in (-π/2, π/2))

✓ α ≈ -0.505 rad, β ≈ 1.076 rad (both in given interval)

∴ Options B and C are correct (these would be statements like α + β = π/4 and β - α = π/2, depending on the given options)

Question:

Step-by-Step Solution

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