If (sin²Ø-3sinØ+2)/(cos²Ø) , where 0° < Ø < 90°, then what is the value of (cos2Ø - sin3Ø+ cosec2Ø)?

If (sin²Ø-3sinØ+2)/(cos²Ø) , where 0° < Ø < 90°…

If $\frac{sin^2Ø-3sinØ+2}{cos^2Ø}$ , where 0° < Ø < 90°, then what is the value of (cos2Ø - sin3Ø+ cosec2Ø)?

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. $\frac{-3+4\sqrt{3}}{6}$

(sin²Ø - 3sinØ + 2)/cos²Ø = 1

(sin²Ø - 3sinØ + 2) = cos²Ø

(sin²Ø - 3sinØ + 2) = 1 - sin²Ø

2 sin²Ø - 3sinØ + 1 = 0

2sin²Ø - 2sinØ - sinØ + 1 = 0

2sinØ((sinØ - 1) - 1(sinØ - 1) = 0

(2sinØ - 1) (sinØ - 1) = 0

sinØ = 1, sinØ = 1/2

Ø = 90⁰ (Not possible), Ø = 30⁰

cos2Ø - sin3Ø + cosec2Ø = cos60⁰ - sin90⁰ + cosec60⁰

= 1/2 - 1 + (2/√3)(√3/√3)

= -1/2 + 2√3/3

= (-3 + 4√3)/6