Let a triangle $A B C$ of sides $a, b, c$, having area $\Delta$


Area $=\Delta$
$\frac{1}{2} b c \sin A=\Delta$ $\ldots(\mathrm{i})$
and $\quad \frac{1}{2} a c \sin B=\Delta$ $\ldots(\mathrm{ii})$
and $\quad \frac{1}{2} a b \sin c=\Delta$ $\ldots(\mathrm{iii})$
Using cosine rule
$a^2=b^2+c^2-2 b c \cos A$
and $\quad b^2=a^2+c^2-2 a c \cos B$
and $\quad c^2=a^2+b^2-2 a b \cos C$
On adding, we get
$a^2+b^2+c^2=2 a^2+2 b^2+2 c^2$ $-2 a b \cos C-2 a c \cos B-2 b c \cos A$
or $a^2+b^2+c^2=2(a b \cos C+a c \cos B+b c \cos A)$ $\ldots(\mathrm{iv})$
Now, from Eq. (i)
$b c=\frac{2 \Delta}{\sin A}$
Eq. (ii), $a c=\frac{2 \Delta}{\sin B}$
Eq. (iii), $\quad a b=\frac{2 \Delta}{\sin C}$
Putting these values in Eq. (iv),
we get $a^2+b^2+c^2=2$ $\left(\frac{2 \Delta}{\sin C} \cos C+\frac{2 \Delta}{\sin B} \cos B+\frac{2 \Delta}{\sin A} \cos A\right)$
$a^2+b^2+c^2=4 \Delta(\cot C+\cot B+\cot A)$
or $\cot C+\cot B+\cot A=\frac{a^2+b^2+c^2}{4 \Delta}$
or $\cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4 \Delta}$