已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).

问题：已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).

答案：↓↓↓

何佩琨的回答：

网友采纳　　(sinβ)^2=(sinα)^2/a^2,(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2.　　(tanβ)^2=(sinα)^2/[a^2-(sinα)^2].　　(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2].　　1/(cosα)^2=b^2/[a^2-(sinα)^2].　　b^2(cosα)^2=a^2-1+(cosα)^2.　　(cosα)^2=(a^2-1)/(b^2-1).