In any `triangle ABC` , if `a^2,b^2,c^2` are in AP then that cot A , cot B , cot C are in are in A.P.

Explanation

By the sine rule ,
`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)=k`
`therefore a=ksin A , b=k sin B , c=k sin C`
Now , `a^2,b^2,c^2` are in A.P.
`therefore c^2-b^2=b^2-a^2`
`therefore b^2=a^2+c^2-b^2`
`therefore (b^2)/(2ac)=(a^2+c^2-b^2)/(2ac)`
`therefore (k^2sin^2B)/(1(ksinA)(ksinC))=cos B`
`therefore (sin^2B)/(2sinA sinC)=cosB`
`therefore (sinB)/(sinA sinC)=(2cosB)/(sinB)`
`therefore (sin[pi-(A+C)])/(sinA sinC)=2cot B`
`therefore (sin(A+C))/(sinA sinC)=2 cot B`
`therefore (sinA cos C +cosA sin C)/(sin A sin C)=2cot B`
`therefore (sin AcosC)/(sin A sinC)+(cosAsinC)/(sinA sinC)=2cot B`
`therefore (cos C)/(sinC)+(cosA)/(sinA)=2cot B`
`therefore cot A+cot C=2 cot B`
Hence , `cot A , cot B , cot C` are in A.P.