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Question

If tanθ=sinαcosαsinα+cosα, then

A
sinαcosα=±2sinθ
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B
sinα+cosα=±2cosθ
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C
cos2θ=sin2α
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D
sin2θ+cos2α=0
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Solution

The correct options are
A cos2θ=sin2α
B sinα+cosα=±2cosθ
C sinαcosα=±2sinθ
D sin2θ+cos2α=0

cos2θ=1sec2θ=11+tan2θ
Using sinαcosαsinα+cosα=tanθ
We get
cos2θ=11+(sinαcosαsinα+cosα)2=(sinα+cosα)22
sinα+cosα=±2cosθ
And
sin2θ=1cos2θ=(sinαcosα)22sinαcosα=±2sinθ
Now,
cos2θ=cos2θsin2θ=(sinα+cosα)22(sinαcosα)22=sin2α
And
sin2θ=2sinθcosθ=cos2α


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