If 3sin2θ + 4cosθ-4 = 0, 0° < θ 90°, then the value of (cosec2θ + cot2θ) is
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Correct answer: A
Explanation
The correct answer is option 1 i.e. 5/4
3sin2θ + 4cosθ - 4 = 0
sin2θ + cos2θ = 1
Hypotenuse(H)2 = Perpendicular(P)2 + Base(B)2
3sin2θ + 4cosθ - 4 = 0
3(1 - cos2θ) + 4cosθ - 4 = 0
3 - 3cos2θ + 4cosθ - 4 = 0
- 3cos2θ + 4cosθ - 1 = 0
3cos2θ - 4cosθ + 1 = 0
3cos2θ - 3cosθ - cosθ + 1 = 0
3cosθ(cosθ - 1) - 1(cosθ - 1) = 0
(cosθ - 1)(3cosθ - 1) = 0
cosθ = 1, 1/3
But cosθ = 1 is not possible because it is given 00 < θ < 900 and cos00 = 1
Thus, cosθ = 1/3
Cosθ = B/H
H2 = B2 + P2
32 = 12 + P2
9 = 1 + P2
9 - 1 = P2
8 = P2
2√2 = P
So, cosec2θ + cot2θ = (3/2√2)2 + (1/2√2)2 = (9/8) + (1/8) = (9 + 1)/8 = 10/8 = 5/4