(5分)若$\tan \theta =-2$,则$\dfrac{\sin \theta (1+\sin 2\theta )}{\sin \theta +\cos \theta }=$( ) A.$-\dfrac{6}{5}$ B.$-\dfrac{2}{5}$ C.$\dfrac{2}{5}$ D.$\dfrac{6}{5}$ 分析:由题意化简所给的三角函数式,然后利用齐次式的特征即可求得三角函数式的值. 解:由题意可得:$\dfrac{\sin \theta (1+\sin 2\theta )}{\sin \theta +\cos \theta }=\dfrac{\sin \theta ({\sin }^{2}\theta +{\cos }^{2}\theta +2\sin \theta \cos \theta )}{\sin \theta +\cos \theta }$ $=\dfrac{\sin \theta {(\sin \theta +\cos \theta )}^{2}}{\sin \theta +\cos \theta }=\sin \theta (\sin \theta +\cos \theta )$ $=\dfrac{{\sin }^{2}\theta +\sin \theta \cos \theta }{{\sin }^{2}\theta +{\cos }^{2}\theta }=\dfrac{{\tan }^{2}\theta +\tan \theta }{1+{\tan }^{2}\theta }$ $=\dfrac{4?2}{1+4}=\dfrac{2}{5}$. 故选:C. 点评:本题主要考查同角三角函数基本关系,三角函数式的求值等知识,属于中等题.
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