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 $\frac{(a+b+c{)}^2}{a^2+b^2+c^2}$


We know, $\cot \frac{A}{2}=\frac{s(s-a)}{\Delta }$
Hence, $\sum \cot \frac{A}{2}=\frac{s^2}{\Delta }\text{and also, we know}\sum \cot A=\frac{a^2+b^2+c^2}{4\Delta }$

Hence, we have:$\frac{\sum \cot \frac{A}{2}}{\sum \cot A}=\frac{(a+b+c{)}^2}{a^2+b^2+c^2}$