Q43 / 160 · TS EAMCET 2003 · Mathematics

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
  • Correct answer
    B
    2
  • C
    3
  • D
    4

Solution

We have $\cot \frac{C}{2} \cot \frac{B}{2}$ $=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \cdot \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\frac{s}{s-a}$ $=\frac{2 s}{2 s-2 a}=\frac{a+b+c}{a+b+c-2 a}$ $=\frac{a+3 a}{3 a-a}=\frac{4 a}{2 a}=2 \quad[\because b+c=3 a]$

Hence, the correct answer is option: 'B'.

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