Trigonometric Equations Question 116
Question: In a $ \Delta ABC, $ if $ 3a=b+c, $ then the value of $ \cot \frac{B}{2}\cot \frac{C}{2} $ is
[Pb. CET 2003; Roorkee 1986; MP PET 1990, 97, 98, 2003; EAMCET 2003; Orissa JEE 2005]
Options:
A) 1
B) 2
C) $ \sqrt{3} $
D) $ \sqrt{2} $
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Answer:
Correct Answer: B
Solution:
- $ \cot \frac{B}{2}.\cot \frac{C}{2}=\sqrt{\frac{s(s-b)}{(s-a),(s-c)}.\frac{s(s-c)}{(s-a),(s-b)}} $ = $ \frac{s}{s-a} $ , {Since $ 3a=b+c $ or $ a+b+c=2s=4a} $ = $ 2a/a=2 $ .
BETA
