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 ........

In any triangle, let the standard notation be
$$a = BC,\; b = CA,\; c = AB$$
and let the area be $$\Delta$$. For the given triangle

$$a = 3,\; b = 4,\; c = \sqrt{23}.$$

Step 1 Find the area using Heron’s formula

Semi-perimeter:
$$s = \frac{a + b + c}{2} = \frac{3 + 4 + \sqrt{23}}{2} = \frac{7 + \sqrt{23}}{2}.$$

Heron’s formula gives
$$\Delta^2 = s(s - a)(s - b)(s - c).$$

Compute each factor:
$$s - a = \frac{7 + \sqrt{23}}{2} - 3 = \frac{1 + \sqrt{23}}{2},$$ $$s - b = \frac{7 + \sqrt{23}}{2} - 4 = \frac{\sqrt{23} - 1}{2},$$ $$s - c = \frac{7 + \sqrt{23}}{2} - \sqrt{23} = \frac{7 - \sqrt{23}}{2}.$$

Hence
$$\Delta^2 = \frac{(7 + \sqrt{23})(1 + \sqrt{23}) (\sqrt{23} - 1)(7 - \sqrt{23})}{16}.$$ Pair-wise products:
$$(7 + \sqrt{23})(7 - \sqrt{23}) = 26,$$ $$(1 + \sqrt{23})(\sqrt{23} - 1) = 22.$$

Therefore
$$\Delta^2 = \frac{26 \times 22}{16} = \frac{143}{4},\qquad \Delta = \frac{\sqrt{143}}{2}.$$

Step 2 Express $$\sin$$ and $$\cos$$ of each angle

Using $$\Delta = \tfrac12 bc \sin A$$, $$\tfrac12 ac \sin B$$, $$\tfrac12 ab \sin C$$:

$$\sin A = \frac{2\Delta}{bc} = \frac{2 \cdot \frac{\sqrt{143}}{2}}{4\sqrt{23}} = \frac{\sqrt{143}}{4\sqrt{23}},$$
$$\sin B = \frac{2\Delta}{ac} = \frac{\sqrt{143}}{3\sqrt{23}},$$
$$\sin C = \frac{2\Delta}{ab} = \frac{\sqrt{143}}{12}.$$

Using the cosine rule:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{16 + 23 - 9}{8\sqrt{23}} = \frac{15}{4\sqrt{23}},$$
$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{9 + 23 - 16}{6\sqrt{23}} = \frac{8}{3\sqrt{23}},$$
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{9 + 16 - 23}{24} = \frac{1}{12}.$$

Step 3 Compute the cotangents

$$\cot A = \frac{\cos A}{\sin A} = \frac{\tfrac{15}{4\sqrt{23}}} {\tfrac{\sqrt{143}}{4\sqrt{23}}} = \frac{15}{\sqrt{143}},$$

$$\cot B = \frac{\cos B}{\sin B} = \frac{\tfrac{8}{3\sqrt{23}}} {\tfrac{\sqrt{143}}{3\sqrt{23}}} = \frac{8}{\sqrt{143}},$$

$$\cot C = \frac{\cos C}{\sin C} = \frac{\tfrac{1}{12}} {\tfrac{\sqrt{143}}{12}} = \frac{1}{\sqrt{143}}.$$

Step 4 Evaluate the required ratio

$$\frac{\cot A + \cot C}{\cot B} = \frac{\dfrac{15}{\sqrt{143}} + \dfrac{1}{\sqrt{143}}} {\dfrac{8}{\sqrt{143}}} = \frac{\dfrac{16}{\sqrt{143}}}{\dfrac{8}{\sqrt{143}}} = \frac{16}{8} = 2.$$

Hence the value of $$\dfrac{\cot A + \cot C}{\cot B}$$ is 2.