A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

$\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$
Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$, $AB = \tan \theta$ and $OB = \sec \theta$. By the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem), \[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] We multiply both sides by $\cos \theta$ to simplify the trigonometric functions, \[AC=OC \sin \theta\] Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$. Therefore, the answer is $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$.