∵sinx=2cosx ∴tanx=sinx/cosx=2
sin?x+1=(sin?x+1)/1=(sin?x+sin?x+cos?x)/(sin?x+cos?x)
分式上下同除cos?x得sin?x+1=(2tan?x+1)/(tan?x+1)=9/5
tanx=2
(sin^2 x+sin^2 x +cos^2 x)/(sin^2 x +cos^2 x)
=同时除以COS^2 X得:
(tan^2 x+tan^2 x+1)/(tan^2 x+1)
带入tanx=2得
原式等于(4+4+1)/(4+1)=9/5
