1.

If `a^2,b^2,c^2`are in A.P., prove that `cotA ,cotB ,cotC`are in `AdotPdot`

Answer» Let `cotA,cotB and cotC` are in `A.P.`
Then, `2cotB = cotA+cotC`
`=>2cosB/sinB = cosA/sinA+cosC/sinC->(1)`
We know,
`sinA/a = sinB/b = sinC/c = k`
`=>sinA = ka, sinB = kb, sinC = kc`
So, (1) becomes,
`2[(a^2+c^2-b^2)/(2ac)]*1/(kb) = [(b^2+c^2-a^2)/(2bc)]*1/(ka) +[(a^2+b^2-c^2)/(2ab)]*1/(kc) `
`=>2a^2+2c^2-2b^2 = b^2+c^2-a^2+a^2+b^2-c^2`
`=>2a^2+2c^2 = 4b^2`
`=>(a^2+c^2)/2 = b^2`
So, `a^2,b^2 and c^2` are in A.P., which is true.
So, our assumption is correct that `cotA,cotB and cotC` are in A.P.


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