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-tga/2*tgb/2
=1
在三角形ABC中,求证:tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=1.
在三角形ABC中,求证:tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=1.
数学人气:880 ℃时间:2020-10-02 02:16:33
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