Cho tam giác ABC. Chứng minh cotA.cotB + cotB.cotC + cotC.cotA = 1.

Ta có:

cotA = -cot(π – A) = -cot(B + C)

=−1tanB+C=−1−tanB.tanCtanB+tanC=tanB.tanC−1tanB+tanC=1−1tanB.tanCtanB+tanCtanB.tanC=1−1tanB.1tanC1tanC+1tanB=1−cotB.cotCcotC+cotB

Suy ra: cotA(cotC + cotB) = 1 – cotB.cotC

Xét cotA.cotB + cotB.cotC + cotC.cotA

= cotA(cotB + cotC) + cotB.cotC

= 1 – cotB.cotC + cotB.cotC

= 1.

Vậy cotA.cotB + cotB.cotC + cotC.cotA = 1.