A circle centred at $O$ has radius $1$ and contains a point $A$ on the circumference. Segment $AB$ is tangent to the circle at $A$ and angle $\measuredangle AOB=\theta$. If point $C$ lies on $OA$ and $BC$ bisects the angle $\measuredangle ABO$, then find $OC$ in terms of $\theta$.
When I solve, I get a quadratic in $OC$, with nasty roots, but the answer is quite simple. Also, I am unable to manipulate my answer to get it. Probably, some easy elegant method.
On
Since, $\Delta OAB$ is right triangle hence, $\angle ABO=90^o-\theta$. Now join the points B & C. As BC is bisector of $\angle ABO$. Hence, $\angle ABC=\angle OBC=\frac{90^o-\theta}{2 }$. Now apply sine rule in $\Delta ABC$ as follows $$\frac{\sin 90^o}{BC}=\frac{\sin \frac{90^o-\theta}{2}}{CA} \tag 1$$
Similarly, apply sine rule in $\Delta OBC$ as follows $$\frac{\sin \theta}{BC}=\frac{\sin \frac{90^o-\theta}{2}}{OC} \tag 1$$ Diving (1) by (2), we get $$\frac{sin 90^o}{\sin \theta}=\frac{OC}{CA}$$$$\implies \frac{CA}{OC}=\sin\theta$$ $$\implies \frac{CA}{OC}+1=\sin\theta +1$$ $$\implies \frac{CA+OC}{OC}=1+\sin\theta$$$$ \implies \frac{AO}{OC}=1+\sin\theta$$ Setting $\color{blue}{AO=\text{radius}=1}$, we get
$$ \implies \frac{1}{OC}=1+\sin\theta$$
$$ \implies \color{blue}{OC=\frac{1}{1+\sin\theta}}$$
Since $OA=1,\angle{AOB}=\theta$, one has $$OB=\frac{1}{\cos\theta},\ \ \ AB=\tan\theta.$$ Hence, $$OB:AB=OC:AC$$ leads $$\tan\theta\times OC=\frac{1}{\cos\theta}\times(1-OC),$$ i.e. $$OC=\frac{1}{1+\sin\theta}.$$