# 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.

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 $$\sin(A)+\sin(B)+\sin(C)=4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\,.$$
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.
• Nice. From your hint, it follows that $$\frac{\cot(A)+\cot(B)}{\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)}=\frac{c}{a+b+c}\,,$$ and so on. However, I suggest that you change the area $A$ to something else, as $A$ is already the angle $A$. – Batominovski Sep 15 '18 at 20:07
• I hope you wouldn't mind me fixing your notations, so $A$ does not represent two different quantities. – Batominovski Sep 15 '18 at 20:10
• Ok in Germany we write $A$ for area and $\alpha,\beta,\gamma$ for the angles of a triangle. – Dr. Sonnhard Graubner Sep 15 '18 at 20:11