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

[asy]import olympiad;
unitsize(2cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
labelmargin=0.2;
dotfactor=3;

pair O=(0,0);
pair A=(1,0);
pair B=(1,1.5);
pair D=bisectorpoint(A,B,O);
pair C=extension(B,D,O,A);

draw(Circle(O,1));
draw(O--A--B--cycle);
draw(B--C);

label("$O$",O,SW);
dot(O);
label("$\theta$",(0.1,0.05),ENE);

dot(C);
label("$C$",C,S);

dot(A);
label("$A$",A,E);

dot(B);
label("$B$",B,E);[/asy]

$ \textbf{(A)}\ \sec^2\theta \minus{} \tan\theta \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\cos^2\theta}{1 \plus{} \sin\theta} \qquad \textbf{(D)}\ \frac {1}{1 \plus{} \sin\theta} \qquad \textbf{(E)}\ \frac {\sin\theta}{\cos^2\theta}$
1. **Understanding the Problem:**
   - We have a circle centered at \( O \) with radius \( 1 \).
   - Point \( A \) is on the circle.
   - Segment \( AB \) is tangent to the circle at \( A \).
   - \( \angle AOB = \theta \).
   - Point \( C \) lies on \( \overline{OA} \) and \( \overline{BC} \) bisects \( \angle ABO \).
   - We need to find \( OC \).

2. **Using the Angle Bisector Theorem:**
   - Let \( \phi \) be the measure of angles \( \angle OBC \) and \( \angle CBA \).
   - By the Angle Bisector Theorem, \( \frac{OB}{OA} = \frac{BC}{CA} \).
   - Since \( OB = 1 \) (radius of the circle) and \( OA = 1 \) (radius of the circle), we have \( \frac{BC}{CA} = 1 \), implying \( BC = CA \).

3. **Using the Law of Sines in \( \triangle OBC \):**
   - In \( \triangle OBC \), we apply the Law of Sines:
     \[
     \frac{OC}{\sin \phi} = \frac{BC}{\sin \theta}
     \]
   - This implies:
     \[
     OC = \frac{\sin \phi}{\sin \theta} BC
     \]

4. **Finding \( \sin \phi \):**
   - In \( \triangle CBA \), we use the fact that \( \sin \phi = \frac{CA}{BC} \).
   - Since \( CA = 1 - OC \) and \( BC = BC \), we have:
     \[
     \sin \phi = \frac{1 - OC}{BC}
     \]

5. **Substituting \( \sin \phi \) into the Law of Sines:**
   - Substituting \( \sin \phi \) back into the formula for \( OC \):
     \[
     OC = \frac{1 - OC}{\sin \theta} BC
     \]
   - Simplifying, we get:
     \[
     OC \sin \theta = 1 - OC
     \]
   - Rearranging terms:
     \[
     OC (1 + \sin \theta) = 1
     \]
   - Solving for \( OC \):
     \[
     OC = \frac{1}{1 + \sin \theta}
     \]

The final answer is \( \boxed{\frac{1}{1 + \sin \theta}} \)