A) 3 :1
B) 2 : 1
C) 1 : 2
D) 1 : 3
Correct Answer: D
Solution :
We have, \[\frac{\tan \text{ }\left( \frac{B}{2} \right)}{\cot \text{ }\left( \frac{C-A}{2} \right)}=\frac{\sin \frac{B}{2}\sin \left( \frac{C-A}{2} \right)}{\cos \frac{B}{2}\cos \text{ }\left( \frac{C-A}{2} \right)}\] \[=\frac{\sin C-\sin A}{\sin C+\sin A}=\frac{kc-ka}{kc+ka}=\frac{c-a}{c+a}=\frac{a}{3a}=\frac{1}{3},\,\{\because \,c=2a\}\].
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