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2021年高考数学甲卷-理9

  2022-05-02 08:58:48  

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.

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