If 7sin2θ + 3cos2θ= 4, 0°

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Home Questions PCS6DMF42R

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Trigonometry

Question:

If 7sin²θ + 3cos²θ= 4, 0° < θ < 90°

than the value of (tan²θ + cosec²2θ) is?

Options:

\(\frac{15}{4}\)

\(\frac{5}{3}\)

\(\frac{13}{4}\)

Correct Answer:

\(\frac{5}{3}\)

Explanation:

θ = 30° satisfying the equation.

So, put θ = 30°

(tan²θ + cosec²2θ)

= (\(\frac{1}{\sqrt {3}}\))² + (\(\frac{2}{\sqrt {3}}\))²

\(\frac{1}{3}\) + \(\frac{4}{3}\) = \(\frac{5}{3}\)