If 4 – 2sin2 θ – 5cos θ = 0, 0°

1.	If 4 – 2sin2 θ – 5cos θ = 0, 0°
Answer» Correct Answer - Option 3 : \(\dfrac{1+2\sqrt{3}}{2}\) Given: 4 – 2sin2 θ – 5cos θ = 0 Identity used: sin²θ = 1 – cos²θ Calculation: 4 – 2sin2 θ – 5cos θ = 0 ⇒ 4 – 2 × (1 – cos2θ) – 5cosθ = 0 ⇒ 4 – 2 + 2cos2θ – 5cosθ = 0 ⇒ 2cos2θ – 5cosθ + 2 = 0 ⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0 ⇒ 2cosθ × (cosθ – 2) – 1 × (cosθ – 2) = 0 ⇒ (2cosθ – 1) × (cosθ – 2) = 0 ⇒ cosθ = 1/2 and cosθ = 2 Rejecting cosθ = 2 as 0° < θ < 90° So, cosθ will be 1/2 ⇒ θ = 60° The value of cos θ + tan θ = cos60° + tan60° ⇒ (1/2) + √3 ∴ The value of cos θ + tan θ is \(\dfrac{1+2\sqrt{3}}{2}\)

Answer» Correct Answer - Option 3 : \(\dfrac{1+2\sqrt{3}}{2}\)

Given:

4 – 2sin2 θ – 5cos θ = 0

Identity used:

sin²θ = 1 – cos²θ

Calculation:

4 – 2sin2 θ – 5cos θ = 0

⇒ 4 – 2 × (1 – cos2θ) – 5cosθ = 0

⇒ 4 – 2 + 2cos2θ – 5cosθ = 0

⇒ 2cos2θ – 5cosθ + 2 = 0

⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ × (cosθ – 2) – 1 × (cosθ – 2) = 0

⇒ (2cosθ – 1) × (cosθ – 2) = 0

⇒ cosθ = 1/2 and cosθ = 2

Rejecting cosθ = 2 as 0° < θ < 90°

So, cosθ will be 1/2

⇒ θ = 60°

The value of cos θ + tan θ = cos60° + tan60°

⇒ (1/2) + √3

∴ The value of cos θ + tan θ is \(\dfrac{1+2\sqrt{3}}{2}\)

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