In $\triangle ABC$, $\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$

Question

In $\Delta ABC$ , prove $$\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$$

I tried to take the $LHS $ in terms of $\sin$ and $\cos$ but ended up getting a huge equation.

I also tried to convert $\cot$ to $\; \dfrac {1}{\tan}\;$ but was still unable to solve the problem.

Answer 1

Hint: Note that $$\begin{align}\frac{\cot(A)+\cot(B)}{\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)}&=\frac{\sin(B)\cos(A)+\sin(A)\cos(B)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\Bigg(\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)+\cos\left(\frac{B}{2}\right)\sin\left(\frac{A}{2}\right)\Bigg)} \\&=\frac{\sin(C)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}\,.\end{align}$$ Similarly, $$\frac{\cot(B)+\cot(C)}{\cot\left(\frac{B}{2}\right)+\cot\left(\frac{C}{2}\right)}=\frac{\sin(A)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}$$ and $$\frac{\cot(C)+\cot(A)}{\cot\left(\frac{C}{2}\right)+\cot\left(\frac{A}{2}\right)}=\frac{\sin(B)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}\,.$$ Thus, the required identity is equivalent to $$\sin(A)+\sin(B)+\sin(C)=4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\,.$$

Answer 2

Hint: Use that $$\cot(\frac{A}{2})=\frac{s}{r_a}$$ etc and $$\cot(A)=\frac{\cos(A)}{\sin(A)}=\frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{2F}{bc}}$$ etc where $$F=\frac{1}{2}bc\sin(A)$$ etc. $$r_a=\frac{F}{s-a}$$ etc.

Answer 3

Note that $$\cot{A} = \frac{b^2+c^2-a^2}{4\Delta};\; \cot{\frac{A}{2}} =\frac{s-a}{r}$$ where $s$ is the semiperimeter, $r$ is the inradius and $\Delta$ is the area of the triangle. Therefore, \begin{align*} \sum_{cyc} {\frac{\cot{A}+\cot{B}}{\cot{\frac{A}{2}}+\cot{\frac{B}{2}}}} &= \sum_{cyc} {\frac{\frac{b^2+c^2-a^2}{4\Delta}+\frac{c^2+a^2-b^2}{4\Delta}}{\frac{s-a}{r}+\frac{s-b}{r}}} \\ &=\sum_{cyc} {\frac{c^2}{2\Delta}\cdot \frac{r}{c}} \\ &= \sum_{cyc} {\frac{c}{2s}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \left[\because \frac{\Delta}{r}=s\right] \\ &= \frac{c+a+b}{a+b+c} =1. \end{align*}

$QED.$