若 tanθ=−2\tan \theta = - 2tanθ=−2 ，则 sinθ(1+sin2θ)sinθ+cosθ=\frac{\sin \theta \left( {1 + \sin {2\theta }}\right) }{\sin \theta + \cos \theta } =sinθ+cosθsinθ(1+sin2θ)= （ ）A. −65- \frac{6}{5}−56B. −25- \frac{2}{5}−52C. 25\frac{2}{5}52D. 65\frac{6}{5}56