In $\triangle A B C$, if $a=3, b=4, c=6$, then $\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot…

In $\triangle A B C$, if $a=3, b=4, c=6$, then $\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=$
  1. $\frac{13}{61}$
  2. $\frac{169}{61}$
  3. $\frac{61}{169}$
  4. $\frac{61}{13}$

Solution

$\begin{aligned} \cot \frac{A}{2}+\cot \frac{B}{2}+\cot & \frac{C}{2}=\frac{s(s-a)}{\Delta} \\ & \quad+\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta} \\ & =\frac{S}{\Delta}[3 S-(a+b+c)] \\ & =\frac{S}{\Delta}(S)=\frac{S^2}{\Delta}=\frac{(a+b+c)^2}{4 \Delta}\end{aligned}$ Now, $ \begin{aligned} & \text { Now, } \cot A+\cot B+\cot C=\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C} \\ &= \frac{b^2+c^2-a^2}{2 b c \sin A}+\frac{c^2+a^2-b^2}{2 a c \sin B}+\frac{a^2+b^2-c^2}{2 a b \sin C} \\ &= \frac{b^2+c^2-a^2}{4 \Delta}+\frac{c^2+a^2-b^2}{4 \Delta}+\frac{a^2+b^2-c^2}{4 \Delta} \\ &= \frac{a^2+b^2+c^2}{4 \Delta} \\ & \frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=\frac{(a+b+c)^2}{a^2+b^2+c^2} \\ &=\frac{(3+4+6)^2}{9+16+36}=\frac{169}{61} . \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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