Value of Ratio of Cotangents in a Given Triangle

In $\triangle ABC$, let the side lengths opposite to the vertices $A, B,$ and $C$ be denoted by $a, b,$ and $c$ respectively.

From the given information: $a = BC = 3$ $b = CA = 4$ $c = AB = \sqrt{23}$

We know that for any triangle $ABC$ with area $\Delta$, the cotangent of an angle can be expressed in terms of its sides as: $\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$ $\cot B = \frac{c^2 + a^2 - b^2}{4\Delta}$ $\cot C = \frac{a^2 + b^2 - c^2}{4\Delta}$

Adding $\cot A$ and $\cot C$, we get: $\cot A + \cot C = \frac{(b^2 + c^2 - a^2) + (a^2 + b^2 - c^2)}{4\Delta} = \frac{2b^2}{4\Delta}$

Now, taking the ratio $\frac{\cot A + \cot C}{\cot B}$: $\frac{\cot A + \cot C}{\cot B} = \frac{\frac{2b^2}{4\Delta}}{\frac{a^2 + c^2 - b^2}{4\Delta}} = \frac{2b^2}{a^2 + c^2 - b^2}$

Substitute the values of $a^2$, $b^2$, and $c^2$: $a^2 = 3^2 = 9$ $b^2 = 4^2 = 16$ $c^2 = (\sqrt{23})^2 = 23$

Substituting these values into the ratio: $\frac{\cot A + \cot C}{\cot B} = \frac{2(16)}{9 + 23 - 16} = \frac{32}{16} = 2$

Thus, the value of the expression is $2$.