Question

In triangle $ABC,\frac{\cot \left(A/2\right)+\cot \left(B/2\right)+\cot \left(C/2\right)}{\cot A+\cot B+\cot C}=$

• A. $\frac{(a+b+c{)}^2}{a^2+b^2+c^2}$
• B. $\frac{a^2+b^2+c^2}{(a+b+c{)}^2}$
• C. $s$
• D. $\Delta$

Answer 1

The correct option is A $\frac{(a+b+c{)}^2}{a^2+b^2+c^2}$

$\Rightarrow \frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}$
$\Rightarrow \frac{\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}}{\frac{b^2+c^2-a^2}{2bc}\times \frac{2R}{a}+\frac{a^2+c^2-b^2}{2ac}\times \frac{2R}{b}+\frac{a^2+b^2-c^2}{2ab}\times \frac{2R}{c}}$
$\Rightarrow \frac{\frac{s(s-a)+s(s-b)+s(s-c)}{\Delta }}{\frac{R}{abc}\times (a^2+b^2+c^2)}$
$(\because \Delta =\frac{abc}{4R})$
$\Rightarrow \frac{4s(s-a+s-b+s-c)}{a^2+b^2+c^2}$
$\Rightarrow \frac{4s(3s-(a+b+c\left)\right)}{a^2+b^2+c^2}$
$\Rightarrow \frac{4s^2}{a^2+b^2+c^2}$
$\Rightarrow \frac{4\times \frac{(a+b+c{)}^2}{4}}{a^2+b^2+c^2}$
$\Rightarrow \frac{(a+b+c{)}^2}{a^2+b^2+c^2}$