在三角形ABC中,求证:tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=1.
人气:425 ℃ 时间:2020-10-02 02:16:33
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-tga/2*tgb/2
=1
推荐
- 在三角形ABC中,求证:tga/2*tgb/2+tgb/2*tgc/2+tgc/2*tga/2=1
- 锐角三角形的三内角为A,B,C,求证tgA·tgB·tgC>1
- 求证:tgA+tgB+tgC=tgAtgBtgC
- 已知tgA=2/3,tgB=4/3,tgC=1/3,求A=?,B=?,C=?
- tgA:tgB:tgC=x:y:z
- 一个扇形的半径是15厘米,面积是141.3平方厘米,这个扇形的圆心角是几度?
- her his him us then their怎么用?
- she was surprised at finding the house empty
猜你喜欢