If \(2{\cos ^2}θ - 5\cos θ + 2 = 0\), 0º

1.	If \(2{\cos ^2}θ - 5\cos θ + 2 = 0\), 0º
Answer» Correct Answer - Option 1 : \(\frac{{\sqrt 3 }}{3}\) Given: 2cos²θ – 5cosθ + 2 = 0 Formula used: Cos60° = 1/2 Cosec60° = 2/(√3) Cot60° = 1/√3 Calculation: 2cos2θ – 5cosθ + 2 = 0 ⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0 ⇒ 2cosθ(cosθ – 2) –1(cosθ – 2) = 0 ⇒ (cosθ – 2)(2cosθ – 1) = 0 ⇒ (cosθ – 2) = 0 or, (2cosθ – 1) = 0 ⇒ cosθ = 2 or, cosθ = 1/2 0º < θ < 90º So, cosθ = 1/2 = cos60° ⇒ θ = 60° 1/(cosecθ + cotθ) = 1/(cosec60° + cot60°) ⇒ 1/[2/(√3) + 1/√3] ⇒ √3/3 ∴ The value of 1/(cosecθ + cotθ) is √3/3.

Answer» Correct Answer - Option 1 : \(\frac{{\sqrt 3 }}{3}\)

Given:

2cos²θ – 5cosθ + 2 = 0

Formula used:

Cos60° = 1/2

Cosec60° = 2/(√3)

Cot60° = 1/√3

Calculation:

2cos2θ – 5cosθ + 2 = 0

⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ(cosθ – 2) –1(cosθ – 2) = 0

⇒ (cosθ – 2)(2cosθ – 1) = 0

⇒ (cosθ – 2) = 0 or, (2cosθ – 1) = 0

⇒ cosθ = 2 or, cosθ = 1/2

0º < θ < 90º

So, cosθ = 1/2 = cos60°

⇒ θ = 60°

1/(cosecθ + cotθ) = 1/(cosec60° + cot60°)

⇒ 1/[2/(√3) + 1/√3]

⇒ √3/3

∴ The value of 1/(cosecθ + cotθ) is √3/3.

Discussion