已知1+tanx/1-tanx=3 求sin2x+2sinx·cosx-cos2x/sin2x+2cos2x已知(1+tanx)/(1-tanx)=3 求(sin^2x+2sinx·cosx-cos^2x)/(sin^2x+2cos^2x)的值
问题描述：
已知1+tanx/1-tanx=3 求sin2x+2sinx·cosx-cos2x/sin2x+2cos2x
已知(1+tanx)/(1-tanx)=3 求(sin^2x+2sinx·cosx-cos^2x)/(sin^2x+2cos^2x)的值
答
数学之美团为你解答(1+tanx) / (1 - tanx) = 3解得 tanx = 1/2(sin²x+2sinxcosx - cos²x) / (sin²x +2cos²x)= (tan²x +2tanx - 1) / (tan²x +2) 【同除以 cos²x 】= (1/4 +1 - 1...