If q is a positive acute angle and 4 cos²θ –

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If q is a positive acute angle and 4 cos²θ – 4 cos θ + 1 = 0, then the value of tan (θ – 15° ) is equal to

1. 0
2. 1
3. √3
4. 1
  √3

Correct Option: B

4cos²θ – 4cosθ + 1 = 0
⇒ (2 cosθ – 1)² = 0
⇒ 2 cosθ – 1 = 0
⇒ 2 cosθ = 1

⇒ cosθ =	1	= cos 60°
	2

⇒ θ = 60°
∴ tan (θ –15°) = tan (60° –15°) =
tan 45° = 1