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1)tan(45°+a)=cosa+sina/cosa-sina
2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y
3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
人气:115 ℃ 时间:2020-04-20 03:32:45
解答
1
tan(45°+a)=1+tana/1-tana=cosa+sina/cosa-sina
2
tan(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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