If 0° < θ < 90°, and tan²θ

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If 0° < θ < 90°, and tan²θ – (√3 + 1) tanθ + √3 = 0, then θ = ?

1. 30°
2. 45°
3. 60°
4. 45° or 60°

Correct Option: D

tan²θ - √3tanθ - tanθ + √3 = 0
= tanθ (tanθ – √3) – 1 (tanθ – √3) = 0
= (tanθ – 1) (tanθ – √3) = 0
= tanθ = 1 = tan 45°
= θ = 45°
Again,
tanθ – √3 = 0
⇒ tanθ = √3 = tan 60°
⇒ θ = 60°.