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Questions  

2sinαcosαsinαcosα is equal to

a
sec⁡α2−π8
b
cos⁡π8−α2
c
tan⁡α2−π8
d
cot⁡α2−π2

detailed solution

Correct option is C

2−sin⁡α−cos⁡αsin⁡α−cos⁡α=2-212sinα+12cosα212sinα-12cosα=2-2cosα-π42sinα-π4=21-cosθ2sinθwhere, θ=α-π4=2sin2θ22sinθ2cosθ2=tanθ2=tanα2-π8

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