In a `Delta ABC`, If tan A/2, tan B/2, tan C/2 are in A.P. then cos A. cos B, cos C are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

A

Explanation

Given tan A/2, tan B/2, tan C/2 are in A.P.
`therefore tan A//2-tan B//2 = tan B//2-tan C//2`
`therefore (sin A//2)/(cos A//2)-(sin B//2)/(cos B//2)=(sin B//2)/(cos B//2)-(sin C//2)/(cos C//2)`
`rArr (sin A//2.cos B//2-sin B//2.cos A//2)/(cos A//2.cos B//2)`
`=(sin B//2.cos C//2-sin C//2.cos B//2)/(cos B//2.cos C//2)`
`rArr (sin((A-B)/(2)))/(cos A//2)=(sin((B-C)/(2)))/(cos C//2)` ...(1)
But `cos A//2 = sin.(B+C)/(2)` and `cos C//2 = sin (A+B)/(2)` ...(2)
`rArr sin((A-B)/(2))sin((A+B)/(2))=sin((B-C)/(2))`
`sin((B+C)/(2))` {from (1) and (2)}
`rArr cos B-cos A=cos C -cos B`
Hence, cos A, cos B, cos C, are in A.P.