题目

已知3cos(2α+β)+5cosβ=0,则tan(α+β)tanα的值为(    )A.±4            B.4                  C.-4             D.1 答案：解析:3cos［(α+β)+α］+5cosβ=0,即3cos(α+β)·cosα-3sin(α+β)·sinα+5cosβ=0.3cos(α+β)cosα-3sin(α+β)sinα+5cos［(α+β)-α］=0,3cos(α+β)cosα-3sin(α+β)·sinα+5cos(α+β)·cosα+5sin(α+β)·sinα=0,8cos(α+β)·cosα+2sin(α+β)·sinα=0,8+2tan(α+β)·tanα=0,∴tan(α+β)-tanα=-4.答案:C--- My mother is ill in hospital.--- ____________.A. I’m sorry to hear that B. I don’ t think soC. It doesn’t matter D. That sounds good