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

Explanation

In `Delta ABC` by sine rule, we have
`a/(sinA) = b/(sin B) = c/(sin C) = k ` …(1)
` :. = k sin A, b = k sin B, c = k sin c`. ….(2)
Consider ` (c-a)/(c+a) = (k sin C - k sin A)/(k sin C + k sin A) ` [From (2) ]
` = (sin C - sin A)/(sin C + sin A) `
`=(2 cos (C+A)/2 * sin (C-A)/2)/(2 sin (C+A)/2 * cos (C-A)/2)`
` = cot ((C+A)/2) * tan ((C-A)/2)`
` = cot(pi/2-B/2) * tan((C-A)/2)" "[:' (A+B+C)/2 = pi/2]`
` :. " " (c-a)/(c+a) = tan. B/2* tan((C-A)/2)`.
` :. " " tan((C-A)/2) = ((c-a)/(c+a)) * cot . B/2 .`