If 4 cos2θ - 3 sin2θ + 2 = 0, then the value of tanθ is (where 0 �

1.	If 4 cos2θ - 3 sin2θ + 2 = 0, then the value of tanθ is (where 0 ≤ θ < 90°) 1. \(\sqrt{2}\)2. \(\sqrt{6}\)3. \(\dfrac{1}{\sqrt{3}}\)4. 1
Answer» Correct Answer - Option 2 : \(\sqrt{6}\) Given 4 cos2 θ - 3 sin2 θ + 2 = 0 Formula: sin2θ + cos2θ = 1 tan2θ = sin2θ/cos2θ Calculation: 4 cos2θ - 3 sin2θ + 2 = 0 ⇒ 4 cos²θ - 3 (1 - cos²θ) + 2 = 0 ⇒ 4 cos²θ - 3 + 3 cos²θ + 2 = 0 ⇒ 7 cos²θ - 1 = 0 ⇒ 7 cos²θ = 1 ⇒ cos²θ = 1/7 sin²θ + cos²θ = 1 ⇒ sin²θ = 1 - 1/7 ⇒ sin²θ = 6/7 Now, tan²θ = sin²θ/cos²θ ⇒ tan²θ = (6/7)/(1/7) ⇒ tan²θ = 6 ∴ tanθ = √6

If 4 cos2θ - 3 sin2θ + 2 = 0, then the value of tanθ is (where 0 ≤ θ < 90°) 1. \(\sqrt{2}\)2. \(\sqrt{6}\)3. \(\dfrac{1}{\sqrt{3}}\)4. 1

Answer» Correct Answer - Option 2 : \(\sqrt{6}\)

Given

4 cos2 θ - 3 sin2 θ + 2 = 0

Formula:

sin2θ + cos2θ = 1

tan2θ = sin2θ/cos2θ

Calculation:

4 cos2θ - 3 sin2θ + 2 = 0

⇒ 4 cos²θ - 3 (1 - cos²θ) + 2 = 0

⇒ 4 cos²θ - 3 + 3 cos²θ + 2 = 0

⇒ 7 cos²θ - 1 = 0

⇒ 7 cos²θ = 1

⇒ cos²θ = 1/7

sin²θ + cos²θ = 1

⇒ sin²θ = 1 - 1/7

⇒ sin²θ = 6/7

Now,

tan²θ = sin²θ/cos²θ

⇒ tan²θ = (6/7)/(1/7)

⇒ tan²θ = 6

∴ tanθ = √6