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Q.

If  cot  θ+tanθ=3,and1cos2θαcosθ=0,then

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a

6α2(α29)=1

b

9α2(α26)=1

c

9α2(6-α2)=1

d

6α2(9α2)=1

answer is C.

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Detailed Solution

cotθ+tanθ=3cos2θ+sin2θsinθcosθ=33sinθcosθ=1.1cos2θαcosθ=0αcosθ=sin2θα3sinθ=sin2θα3=sin3θ.αcosθ=19cos2θcos3θ=19α.sin2θ+cos2θ=1α32/3+19α2/3=119α2/3=1α32/319α2=1α32/33=1α323α32/319α2/3=1α2913=23α29181α2=6α299α2(6α2)=1.

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