高一数学求证下列恒等式

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1 | 2012-7-25 12:32:14 | 显示全部楼层 |阅读模式

1tan(45°+a)=1+tana/1-tana=cosa+sina/cosa-sina2tan(x+y)tan(x-y)=[(tanx+tany)(tanx-tany)]/[(1-tanxtany)(1+tanxytany)]=(tan2x-tan2y)/(1-tan2xtan2y)3.tanx+tany/tanx-tany=[(sinx/cosx)+(siny/cosy)]/[(sinx/cosx)-(siny/cosy)]=(sinxcosy+sinycosx)/(sinxcosy-cosxsiny)=sin(x+y)/sin(x-y)...

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