Trigonometric Equations Question 180
Question: If in the $ \Delta ABC,AB=2BC $ , then $ \tan \frac{B}{2}:\cot ( \frac{C-A}{2} ) $
Options:
A) 3 :1
B) 2 : 1
C) 1 : 2
D) 1 : 3
Show Answer
Answer:
Correct Answer: D
Solution:
- We have, $ \frac{\tan ( \frac{B}{2} )}{\cot ( \frac{C-A}{2} )}=\frac{\sin \frac{B}{2}\sin ( \frac{C-A}{2} )}{\cos \frac{B}{2}\cos ( \frac{C-A}{2} )} $ $ =\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} $ .
BETA
