求证tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
热心网友
tan(x-y)+tan(y-z)+tan(z-x)=tan[(x-y)+(y-z)][1-tan(x-y)tan(y-z)]+tan(z-x)=tan(x-z)[1-tan(x-y)tan(y-z)]+tan(z-x)=tan(x-z)-tan(x-z)tan(x-y)tan(y-z)+tan(z-x)=-tan(z-x)-tan(x-z)tan(x-y)tan(y-z)+tan(z-x)=-tan(x-z)tan(x-y)tan(y-z)]=tan(z-x)tan(x-y)tan(y-z)]=右边证毕
热心网友
咳！∵（x-y）+（y-z）+（z-x）=0∴(x-y)+(y-z)=-(z-x)两边同时取正切得：tan[(x-y)+(y-z)]=-tan（z-x）tan(x-y)+tan(y-z)/1-tan(x-y)tan(y-z)=-tan(z-x)tan(x-y)+tan(y-z)=-tan(z-x)+tan(x-y)tan(y-z)tan(z-x)∴tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)∴原式成立。