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# if sides of a tringle  a,b,c are in AP then prove that :COSA.COTA/2, COSB.COTB/2, COSC.COTC/2 ARE IN AP.

ANIL KUMAR
10 Points
3 years ago
Since a,b,c are in AP hence 2b=(a+c)...................................(1)
sinA/a=sinB/b=sinC/c=k or we can say that sinA=ka,sinB=kb,sinC=kc hence
sinA, sinB, sinC are also in AP[since a,b,c are in AP hence multiplyimg by constant ka,kb,kc are also in AP]
hence sinB=1/2[sinA+sinC].............................................................(2)

Now in Question Asume that cosA.cotA/2,cosB.cotB/2,cosC.cotC/2 are in AP
hence:
2cosB.cotB/2=cosA.cotA/2 +cosC.cotC/2---------------------------------------(3)
LHS                                  RHS
take LHS:
= 2cosB.cotB/2
=2cotB/2.(1-sin^2B/2)
=2cotB/2-2sinB/2.cosB/2
=2cotB/2-sinB........................................................(4)
RHS:
=cosA.cotA/2+cosC.cotC/2
=(1-sin^2(A/2))cotA/2+(1-sin^2(C/2))cotC/2
=cotA/2+cotC/2-1/2[2sinA/2.cosA/2+2sinC/2.cosC/2]
=cotA/2+cotC/2-1/2[sinA+sinC]
=cotA/2+cotC/2-sinB                    [using (2)].....................................(5)
=2cotB/2-sinB       by using[6]
=LHS

Now using
by cotangent principle if s=(a+b+c)/2 and r is redius circcle of inside tringle: then
(cotA/2)/(s-a)=(cotB/2)/(s-b)=(cotC/2)/(s-c)=r
by this we get:
cotA/2+cotC/2
=(s-a)/(s-b).cotB/2 +(s-c)/(s-b)cotB/2
=cotB/2[1/(s-b).(2s-(a+c))]
=cotB/2.[(2s-2b)/(s-b)]....................by(1) a+c=2b
=2cotB/2
=cotB/2[]..........................................................(6)...........now use itin

Friends I solved it in 15 minutes after getting question from my friend who is lecturer ic an Inter College: I am working Assistent Professor in PSIT Kanpur and did BTech and MTech in CSE discipline.