求证：tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
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解决时间 2021-08-21 11:29
- 提问者网友：聂風
- 2021-08-20 11:59
求证：tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
最佳答案
- 五星知识达人网友：青灯有味
- 2021-08-20 12:37
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
写得很清楚的,你难道还看不懂么?
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
写得很清楚的,你难道还看不懂么?
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