10．（2022秋•宁波期末）已知tanα=3\tan \alpha =3tanα=3，则sin(π−α)+2cos(π+α)sin(π2+α)+cos(3π2+α)=(\frac{\sin ({\pi -\alpha })+2\cos ({\pi +\alpha })}{\sin ({\frac{\pi }{2}+\alpha })+\cos ({\frac{3\pi }{2}+\alpha })}=(sin(2π+α)+cos(23π+α)sin(π−α)+2cos(π+α)=( )))