If 3 tan θ = 2√3 sinθ, 0º (cosec² 2 θ + cot² 2θ)/(sin²θ+tan² 2θ) is:

If 3 tan θ = 2√3 sinθ, 0º (cosec² 2 θ + cot²…

If 3 tan θ = 2√3 sinθ, 0º < θ < 90º, then the value of $\frac{cosec^2 2 \theta + cot^2 2\theta}{sin^2\theta+tan^2 2\theta}$ is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. $\frac{20}{39}$

3 tan θ = 2√3 sinθ

cosθ = √3/2 = cos30

θ = 30°.

= $\frac{cosec^2 2 \theta + cot^2 2\theta}{sin^2\theta+tan^2 2\theta}$

= [cosec²(60) + cot²(60)]/[sin² 30 + tan²60]

= [4/3 + 1/3]/[1/4 + 3]

= (5/3)/(13/4)

= 20/39.