If 2 sinθ – 8 cos²θ + 5 = 0, 0ᵒ < θ < 90ᵒ, then what is the value of (tan 2θ + cosec 2θ) ?

If 2 sinθ – 8 cos²θ + 5 = 0, 0ᵒ < θ < 90ᵒ, then what is…

If 2 sinθ – 8 cos²θ + 5 = 0, 0^o < θ < 90^o, then what is the value of (tan 2θ + cosec 2θ) ?

Show answer & explanation

Correct answer: A

Explanation

The correct answer is Option 1 i.e 5√(3)/3.

Understanding/Application

2sinθ - 8cos²θ + 5 = 0

2sinθ - 8(1 - sin²θ) + 5 = 0

8sin²θ + 2sinθ - 3 = 0

8sin²θ+ 6sinθ - 4sinθ - 3 = 0

2sinθ(4sinθ + 3) -1(4sinθ + 3) = 0

(2sinθ - 1) (4sinθ + 3) = 0

sinθ = 1/2, -3/4

-3/4 not possible since θ ∈ (0, 90)

θ = π/6 = 30⁰

Thus, tan2θ + cosec2θ

= tan60⁰ + cosec60⁰

= ${\sqrt{3} }$ + 2/ ${\sqrt{3} }$

= 5/ ${ \sqrt{3}}$

= 5 ${\sqrt{3} }$ /3