If 7Cos²θ + 3Sin²θ = 6, 0ᵒ < θ < 90ᵒ, then the value of (Cot²2θ + Sec²2θ)/(tan²2θ - Sin²2θ) is:

If 7Cos²θ + 3Sin²θ = 6, 0ᵒ < θ < 90ᵒ, then the value of…

If 7Cos²θ + 3Sin²θ = 6, 0^o < θ < 90^o, then the value of (Cot²2θ + Sec²2θ)/(tan²2θ - Sin²2θ) is:

Show answer & explanation

Correct answer: C

Explanation

The correct answer is Option 3 i.e 52/27.

Understanding/Application

7Cos²θ + 3Sin²θ = 6

7(1 - Sin²θ) + 3Sin²θ = 6

7 - 7Sin²θ + 3Sin²θ = 6

4Sin²θ = 1

Sinθ = 1/2

For, 0⁰ < θ < 90⁰

θ = 30⁰

(Cot²2θ + Sec²2θ)/(tan²2θ - Sin²2θ)

= (Cot²2(30) + Sec²2(30))/(tan²2(30) - Sin²2(30))

= (Cot²60⁰ + Sec²60⁰)/(tan²60⁰ - Sin²60⁰)

= {(1/√3)² + 2²}/{(√3)² - (√3/2)²}

= (1/3 + 4)/(3 - 3/4)

= (13/3)/(9/4)

= 52/27