A circle centered at O has radius 1 and contains point ,A Segment AB is tangent to the circle at A and ∠AOB = θIf point C lies on OA, and BC bisects the angle ABO, then OC equals

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sec⁡θ(sec⁡θ−tan⁡θ)

cos2⁡θ1+sin⁡θ

11+sin⁡θ

1−sin⁡θcos2⁡θ

answer is A.

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Detailed Solution

Using property of angle bisector, we get OBAB=OCAC⇒sec⁡θtan⁡θ=x1−x or x=sec⁡θsec⁡θ+tan⁡θ =sec⁡θ(sec⁡θ−tan⁡θ)=11+sin⁡θ

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