If tanθ + 3cotθ - 2√3 = 0, 0° < θ < 90°, then what is the value of (cosec2θ + cos2θ)?
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Correct answer: B
Explanation
The correct answer is option 2 i.e.
tanθ + 3cotθ - 2√3 = 0,
sinθ/cosθ + 3cosθ/sinθ = 2√3
sin2θ + 3cos2θ = 2√3sinθcosθ
(sinθ - √3cosθ)2 = 0
sinθ = √3cosθ
tanθ = tan60
θ = 60.
cosec2θ + cos2θ = cosec260 + cos260
= 4/3 + 1/4 = 19/12.