求证,tan+tan+tan=tantantan
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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
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