作业帮求证,tan+tan+tan=tantantan
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/28 08:29:11
求证,tan+tan+tan=tantantan
两角和正切公式：
tan[(x-y)+(y-z)]=[tan(x-y)+tan(y-z)]/[1-tan(x-y)tan(y-z)]
tan(x-y)+tan(y-z)
=tan(x-y+y-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)-tan(x-z)tan(x-y)tan(y-z)
=-tan(z-x)+tan(z-x)tan(x-y)tan(y-z)
所以tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)成立
tan[(x-y)+(y-z)]=[tan(x-y)+tan(y-z)]/[1-tan(x-y)tan(y-z)]
tan(x-y)+tan(y-z)
=tan(x-y+y-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)*[1-tan(x-y)tan(y-z)]
=tan(x-z)-tan(x-z)tan(x-y)tan(y-z)
=-tan(z-x)+tan(z-x)tan(x-y)tan(y-z)
所以tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)成立
tan
tanα+tanβ+tan(α+β)tanαtanβ=tan(α+β)求证左边等于右边
求证tanB=3tan
高一数学 急求证：tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)
证明:tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
1.求证tanαsinα/tanα-sinα=tanα+sinα/tanαsinα
求证:tan(nπ+a)=tana
求证tan(360°-a)=-tana
cosa=cosb*cosr求证tan（a+b）*tan（a-b)=tan^2（r/2)
已知2tanα=3tanβ.求证：tan(α-β)=sin2β/(5-cos2β)
证明tan阿尔法+tan贝塔=tan(阿尔法+贝塔)-tan阿尔法tan贝塔tan(阿尔法+贝塔)要详解
证明 tan(nA/2)tan(nB/2)+tan(nB/2)tan(nC/2)+tan(nA/2)tan(nC/2)=