In `triangle ABC` tan(A/2), tan(B/2) and tan (C/2)` are in A.P. then prove that cosA,cosB,cosC are in A.P.

Explanation

`tanB/2-tanA/2=tanC/2-tanB/2`

`(sinB/2)/(cosB/2)-(sinA/2)/(cosA/2)=(sinC/2)/(cosC/2)-(sinB/2)/(cosB/2)`

`(sinB/2*cosA/2-sinA/2*cosB/2)/(cosB/2*cosA/2)=(sinC/2*cosb/2-sinB/2*cosC/2)/(cosC/2*cosB/2)`

`(sin((B-A)/2))/(cos(A/2))=(sin((C-B)/2))/(cos(C/2))`

`(-sin((A-B)/2))/(cos(pi/2-(B+C)/2))=9-sin((B-C)/2)/(cos(pi/2-(A+B)/2)`

`2sin((A-B)/2)sin((A+B)/2)=2sin((B-C)/2)sin((B+C)/2)`

`cosB-cosA=cosC-cosB`

`cosAcosBcosC`.