Ans: B
For $0^\circ \le \theta \le 360^\circ$, we have
For $0^\circ \le \theta \le 360^\circ$, we have
$\begin{array}{rcccl}
-1 & \le & \sin \theta & \le & 1 \\
1 & \le & 2 + \sin \theta & \le & 3
\end{array}$
and
$\begin{array}{rcccl}
-1 & \le & \sin \theta & \le & 1 \\
-1 & \le & -\sin \theta & \le & 1 \\
1 & \le & 2 -\sin \theta & \le & 3
\end{array}$
Hence, we have
$\begin{array}{rcccl}
\dfrac{1}{3} & \le & \dfrac{2+\sin\theta}{2-\sin\theta} & \le & \dfrac{3}{1} \\
\dfrac{1}{3} & \le & \dfrac{2+\sin\theta}{2-\sin\theta} & \le & 3 \\
\end{array}$
Therefore, the least value is $\dfrac{1}{3}$.
