In any triangle `A B C ,`prove that:`(b-c)cot A/2+(c-a)cot B/2+(a-b)co

1.	In any triangle `A B C ,`prove that:`(b-c)cot A/2+(c-a)cot B/2+(a-b)cot C/2`=0
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"` `rArr" "a=ksinA, b=ksin B and C=sinC.` `:." "(b-c)cot""A/2=k(sinB-sinC)cot""A/2` `=2kcos""((B+C))/2sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2kcos(pi/2-A/2)sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2ksin""A/2sin""((B-C))/2(cos(A/2))/(sin(A/2))` `=2ksin""((B-C))/2cos""A/2` `=2ksin((B-C))/2cos{pi/2-((B+C))/2}` `=2ksin""((B-C))/2sin""((B+C))/2` `=k(cosC-cosB)`. Similarly, we have `(c-a)cot""B/2=k(cosA-cosC)and (a-b)cos""C/2=k(cosB-cosA).` `:." "(b-c)cos""A/2+(c+a)cos""B/2+(a-b)cot""C/2` `=k[(cosC-cosB)+(cosA-cosC)+(cosB-cosA)]=0`.

Discussion