Question

In a traingle ABC, prove that Tan(C-A)/2 = (c-a)(c+a) cotB/2.​

Answer 1

Step-by-step explanation:

In ∆ABC,by sine Rule

a/sinA=b/sinB=c/sinC=K

a=K sinA, b=K sinB, c=K sinC

Consider,

c-a/c+a=K sinC-K sinA/K sinC+K sinA

=sinC-sinA/sinC+sinA

=2cos[(C+A)/2] sin[(C-A)/2]

/ 2sin[(C+A)/2]cos[(C-A)/2]

=cot[(C+A)/2]tan[(C-A)/2]

=tanB/2 tan(C-A)/2

So, tan(C-A)/2=(c-a) (c+a) cot B/2 (Proved)