证明(cotA+cotB)(cotB+cotC)(cotC+cotA)=cscAcscBcscC

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证明(cotA+cotB)(cotB+cotC)(cotC+cotA)=cscAcscBcscC
证明(cotA+cotB)(cotB+cotC)(cotC+cotA)=cscAcscBcscC

证明(cotA+cotB)(cotB+cotC)(cotC+cotA)=cscAcscBcscC
(cotA+cotB)(cotB+cotC)(cotC+cotA)
=(cosA/sinA+cosB/sinB)(cosB/sinB+cosC/sinC)(cosC/sinC+cosA/sinA)
=(cosAsinB+sinAcosB)/sinAsinB*(cosBsinC+sinBcosC)/sinBsinC*(cosCsinA+sinCcosA)/sinCsinA
=sin(A+B)/sinAsinB*sin(B+C)/sinBsinC*sin(A+C)/sinCsinA
=sinC/sinAsinB*sinA/sinBsinC*sinB/sinCsinA
=1/sinAsinBsinC
=cscAcscBcscC.
前提是在△ABC中!