In $\triangle ABC$, $\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$ 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.
 A: 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
$$\sin(A)+\sin(B)+\sin(C)=4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\,.$$
A: 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.
