已知(1+tanx)/(1-tanx)=3+2√2,求(sin^2x+√2sinx*cosx-cos^2x)/(sin2x+cos^x)的已知(1+tanx)/(1-tanx)=3+2√2,求（sin^2x+√2sinx*cosx-cos^2x）/（sin2x+cos^x）的值
问题描述：
已知(1+tanx)/(1-tanx)=3+2√2,求(sin^2x+√2sinx*cosx-cos^2x)/(sin2x+cos^x)的
已知(1+tanx)/(1-tanx)=3+2√2,求（sin^2x+√2sinx*cosx-cos^2x）/（sin2x+cos^x）的值
答
(sin^2 x+√2sinx*cosx-cos^2 x)/(sin^2 x+2cos^2 x)
=(tan^2x+√2tanx-1)/(tan^2x+2)
=(0.5+1-1)/(0.5+2)
=0.2
答
（1+tanx）/(1-tanx) =3+2√2 ,
解得tanx=√2/2.
(sin^2 x+√2sinx*cosx-cos^2 x)/(sin^2 x+2cos^2 x)
=(tan^2x+√2tanx-1)/(tan^2x+2)
=(1/2+1-1)/(1/2+2)
=1/5.