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

$\frac{\cot A+\cot C}{\cot B}=\frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}$

$=\frac{\cos A\sin C+\cos C\sin A}{\sin A\sin C}·\frac{\sin B}{\cos B}$

$=\frac{\sin (A+C)}{\sin A\sin C}·\frac{\sin B}{\cos B}$

$=\frac{\sin (\pi -B)}{\sin A\sin C}·\frac{\sin B}{\cos B}$

$=\frac{{\sin }^{2}B}{\sin A\sin C\cos B}$

Let, $AB=c, BC=a, CA=b$

By using sine rule, we get

$=\frac{b^2}{ac\cos B}$

$=\frac{b^2}{ac\left(\frac{a^2+c^2-b^2}{2ac}\right)}$

$=\frac{2b^2}{a^2+c^2-b^2}$

$=\frac{32}{9+23-16}$

$=\frac{32}{16}=2$