1986 AHSME Problem 21

In the configuration below, \theta is measured in radians, C is the center of the circle, BCD and ACE are line segments and AB is tangent to the circle at A.

A necessary and sufficient condition for the equality of the two shaded areas, given 0 < \theta < \frac{\pi}{2}, is

\begin{array}{l} \textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad \textbf{(C)}\ \tan\theta = 4\theta\qquad \textbf{(D)}\ \tan 2\theta =\theta\qquad\\ \textbf{(E)}\ \tan\frac{\theta}{2}=\theta \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.