骑行双肩包品牌:求证tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-a)tan(z-x)

来源:百度文库 编辑:神马品牌网 时间:2024/05/07 04:32:34

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)

tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)