On the image below line DB is tangent to the circle. The problem is to try and express both $\Delta OBC$ and $\Delta OBD$ in terms of $\sin\theta$ and $\cos\theta$.
For $\Delta OBC$ $$A=\frac{1}{2}OB(CA)$$ I found out all the trig functions for it$$\sin\theta=\frac{CA}{OC}$$ $$\cos\theta=\frac{OA}{OC}$$ $$\tan\theta=\frac{CA}{OA}$$ I am having trouble using the trig functions to express area in $\sin\theta$ and $\cos\theta$
$\Delta OBC$ and $\Delta OBD$ are not similar triangles. $\Delta OBC$ is not a right triangle.
Also, by "$\theta$", I assume you mean $x$? If so, $\tan\theta\neq\frac{CA}{OC}$, it is $\tan\theta=\frac{CA}{OA}$. Because $OC = 1$, the trig functions simplify to $$\sin \theta = CA $$ $$\cos \theta = OA$$
$$A_{OBC}= \frac{1}{2}(OB)(CA) = \frac{1}{2}(1)(\sin \theta) = \frac{1}{2}\sin \theta $$