In a triangle ABC, let $\mathrm{AB}=\sqrt{23},\mathrm{BC}=3$ and CA = 4. Then the value of $\frac{\cot \mathrm{A}+\cot \mathrm{C}}{\cot \mathrm{B}}$ is

$\mathrm{AB}=\sqrt{23},\mathrm{BC}=3,\mathrm{CA}=4$

$\cos \mathrm{A}=\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}$

$\frac{\cot \mathrm{A}+\cot \mathrm{C}}{\cot \mathrm{B}}$$=\frac{\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bcsinA}}+\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{ab}\cdot \sin \mathrm{C}}}{\frac{{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2}{2\mathrm{ac}\cdot \sin \mathrm{B}}}$

$\text{=}\frac{\frac{b^2+c^2-a^2}{4\mathrm{\Delta }}+\frac{a^2+b^2-c^2}{4\mathrm{\Delta }}}{\frac{c^2+a^2-b^2}{4\mathrm{\Delta }}}$

$\begin{array}{l}\frac{\cot \mathrm{A}+\cot \mathrm{C}}{\cot \mathrm{B}}=\frac{2{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{c}}^2-b^2}=\frac{2\left(16\right)}{9+23-16}=2\end{array}$