$a \cot A + b \cot B + c \cot C = 2(R+ r)$

Question

With usual notations, prove that in a triangle $ABC$:

$$a \cot A + b \cot B + c \cot C = 2(R+ r)$$

I know $R = (abc)/(4\Delta)$ and $r = \Delta/s$.

I could not understand how to start it.

Answer 1

\begin{align} a \cot A + b \cot B + c \cot C &= 2R\Bigg[\sin A \frac{\cos A}{\sin A} + \sin B \frac{\cos B}{\sin B} + \sin C \frac{\cos C}{\sin C}\Bigg]\\ &= 2R\big[\cos A + \cos B +\cos C\big]\\ &= 2R\Bigg[2\cos \frac{A+B}{2}\cos \frac{A-B}{2} + 1-2\sin^2 \frac{C}{2}\Bigg]\\ &=2R\Bigg[1+ 2\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2}\Bigg]\\ &=2R\Big[1+\frac{r}{R}\Big] =2(R+r). \end{align} Hope it helps.

Answer 2

$\sum a\cot A = \sum \dfrac{a}{\sin A}\cos A = \sum 2R\cos A = 2R(1+\dfrac{r}{R}) = 2(r+R)$, where $\sum\cos A = 1+\dfrac{r}{R}$, is found here.