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

\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.
