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Question

3tanθ+cotθ=5cosec θ.
Solve for θ, 0 ≤ θ ≤ 90.

A

60°

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B
45°
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C
150°
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D
30°
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Solution

The correct option is A

60°


3sinθcosθ +cosθsinθ =5sinθ

sinθ (3sin2θ+cos2θ)=5(sinθ cosθ)

3(1cos2θ)+cos2θ=5cosθ
2cos2θ+5cosθ3=0
2cosθ[cosθ+3]1(cosθ+3)=0
(cosθ+3)(2cosθ1)=0

cosθ=3 (Not possible as cosθ ≤ 1)
cosθ=12
θ=60


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