TS EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C\), if \(3 a=b+c\), then \(\cot \frac{B}{2} \cot \frac{C}{2}\) is equal to :
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
We have \(\cot \frac{C}{2} \cot \frac{B}{2}\) \(=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \cdot \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\frac{s}{s-a}\) \(=\frac{2 s}{2 s-2 a}=\frac{a+b+c}{a+b+c-2 a}\) \(=\frac{a+3 a}{3 a-a}=\frac{4 a}{2 a}=2 \quad[\because b+c=3 a]\)
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