已知1+tanx/1-tanx=3 求sin2x+2sinx·cosx-cos2x/sin2x+2cos2x
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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
已知(1+tanx)/(1-tanx)=3 求(sin^2x+2sinx·cosx-cos^2x)/(sin^2x+2cos^2x)的值
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(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) / (1/4 +2)
= 1/9
(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) / (1/4 +2)
= 1/9
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