Question

In `triangle ABC` prove that `tan((C-A)/2) = ((c-a)/(c+a))cot  B/2`

Answer 1

In `triangle ABC`, by sine Rule

`a/(sin A) = b/(sin B) = c/(sin C) = K`

∴ a =  k sinA, b = k sinB, c = k sinC

Consider,

`(c-a)/(c+a) = (ksinC - ksinA)/(ksinC + ksin A)`

`= (sinC - sin A)/(sin C + sin A)`

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

`= cot((C+A)/2)tan((C-A)/2)`

`= tan  B/2 tan((C-A)/2)`

`:. tan((C-A)/2) = ((C-a)/(C+a)) cot  B/2`

Hence proved