Question

Let λ₁λ₂ ∈[0, π] are the solutions of the equation `"cosec"(π/4 + x) + "cosec"(π/4 - x) = 2sqrt(2)`, then 8(sin²λ₁ + sin²λ₂) is equal to ______.

विकल्प

• 3.00

• 4.00

• 5.00

• 6.00

Answer 1

Let λ₁λ₂ ∈[0, π] are the solutions of the equation `"cosec"(π/4 + x) + "cosec"(π/4 - x) = 2sqrt(2)`, then 8(sin²λ₁ + sin²λ₂) is equal to 6.00.

Explanation:

`1/(sin(π/4 + x)) + 1/(sin(π/4 - x)) = 2sqrt(2)`

`sqrt(2)/(cosx + sinx) + sqrt(2)/((cosx - sinx)) = 2sqrt(2)`

`cosx/(cos^2x - sin^2x)` = 1

cosx = 2cos²x – 1

2cos²x – cosx – 1 = 0

2cos²x – 2cosx + cosx – 1 = 0

cosx = 1, cosx = `-1/2`

λ₁ = 0, λ₂ = `(2π)/3`

sin²λ₁ +  sin²λ₂ = `0 + 3/4`

sin²λ₁ +  sin²λ₂ = `3/4`

8(sin²λ₁ +  sin²λ₂)

`8 xx 3/4` = 6