在三角形ABC中,求证：tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=1
人气：186 ℃ 时间：2020-10-02 02:16:32
解答
tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=tga/2*tgb/2+tgc/2*(tga/2+tgb/2)=tga/2*tgb/2+tg[90-(a+b)/2]*(tga/2+tgb/2)=tga/2*tgb/2+ct(a/2+b/2)*(tga/2+tgb/2)=tga/2*tgb/2+(tga/2+tgb/2)/tg(a/2+b/2)=tga/2*tgb/2+1-t...
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