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

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$\sec θ (\sec θ - \tan θ)$

$\frac{1}{1 + \sin θ}$

$\frac{\cos^{2} θ}{1 + \sin θ}$

$\frac{1 - \sin θ}{\cos^{2} θ}$

answer is A, C, D.

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Detailed Solution

Using property of angle bisector, we get

$\begin{array}{l} \frac{OB}{AB} = \frac{OC}{AC} \Rightarrow \frac{\sec θ}{\tan θ} = \frac{x}{1 - x} \\ or \begin{aligned} x & = \frac{\sec θ}{\sec θ + \tan θ} \\ = \sec θ (\sec θ - \tan θ) = \frac{1}{1 + \sin θ} \end{aligned} \end{array}$

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