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If 4 cos2θ - 3 sin2θ + 2 = 0, then the value of tanθ is (where 0 ≤ θ < 90°)

A2\sqrt{2}

B6\sqrt{6}

C13\frac{1}{\sqrt{3}}

D1

Answer:

6\sqrt{6}

Read Explanation:

Solution:

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


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