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Question

In ∆ABC, if a2, b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.

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Solution

Let sinAa=sinBb=sinCc=k

Then,
sinA=ka, sinB=kb, sinC=kc
a2, b2 and c2 are in A.P.
2b2=a2+c2 2a2+c2-b2=22b2-b2=2b2=b2+b2+c2-a2-c2+a22a2+c2-b2=b2+c2-a2+a2+b2-c22a2+c2-b22abc=b2+c2-a22abc+a2+b2-c22abc2cosBkb=cosAka+cosCkc2cosBsinB=cosAsinA+cosCsinC2cotB=cotA+cotCcotA,cotB and cotC are in AP.

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