9．（5分）若$\alpha \in (0,\dfrac{\pi }{2})$，$\tan 2\alpha =\dfrac{\cos \alpha }{2-\sin \alpha }$，则$\tan \alpha =$(　　)
A．$\dfrac{\sqrt{15}}{15}$              B．$\dfrac{\sqrt{5}}{5}$              C．$\dfrac{\sqrt{5}}{3}$              D．$\dfrac{\sqrt{15}}{3}$
分析：把等式左边化切为弦，再展开倍角公式，求解$\sin \alpha$，进一步求得$\cos \alpha$，再由商的关系可得$\tan \alpha$的值．
解：由$\tan 2\alpha =\dfrac{\cos \alpha }{2-\sin \alpha }$，得$\dfrac{\sin 2\alpha }{\cos 2\alpha }=\dfrac{\cos \alpha }{2-\sin \alpha }$，
即$\dfrac{2\sin \alpha \cos \alpha }{1-2si{n}^{2}\alpha }=\dfrac{\cos \alpha }{2-\sin \alpha }$，
$\because \alpha \in (0,\dfrac{\pi }{2})$，$\therefore \cos \alpha \ne 0$，
则$2\sin \alpha (2-\sin \alpha )=1-2\sin ^{2}\alpha$，解得$\sin \alpha =\dfrac{1}{4}$，
则$\cos \alpha =\sqrt{1-si{n}^{2}\alpha }=\dfrac{\sqrt{15}}{4}$，
$\therefore \tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }=\dfrac{\dfrac{1}{4}}{\dfrac{\sqrt{15}}{4}}=\dfrac{\sqrt{15}}{15}$．
故选：$A$．
点评：本题考查三角函数的恒等变换与化简求值，考查倍角公式的应用，是基础题．