The value of $$\cot \frac{\pi}{24}$$ is:

We have to evaluate $$\cot\dfrac{\pi}{24}$$. Since $$\dfrac{\pi}{24}=7.5^{\circ}$$ and $$7.5^{\circ}=\dfrac{15^{\circ}}{2}$$, it is convenient to use the half-angle identity for cotangent.

The cotangent half-angle identity is stated as

$$\cot\dfrac{\theta}{2}\;=\;\dfrac{1+\cos\theta}{\sin\theta}\,.$$

Putting $$\theta=15^{\circ}$$ gives

$$\cot7.5^{\circ}\;=\;\dfrac{1+\cos15^{\circ}}{\sin15^{\circ}}.$$

Now we need the exact values of $$\sin15^{\circ}$$ and $$\cos15^{\circ}$$. We obtain them from the compound-angle formulas with $$15^{\circ}=45^{\circ}-30^{\circ}$$.

The sine difference formula is

$$\sin(A-B)=\sin A\cos B-\cos A\sin B.$$

Hence

$$\sin15^{\circ}= \sin(45^{\circ}-30^{\circ}) = \sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}.$$

Substituting the exact values $$\sin45^{\circ}=\dfrac{\sqrt2}{2},\; \cos45^{\circ}=\dfrac{\sqrt2}{2},\; \cos30^{\circ}=\dfrac{\sqrt3}{2},\; \sin30^{\circ}=\dfrac12,$$ we get

$$\sin15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}-\dfrac{\sqrt2}{4} =\dfrac{\sqrt6-\sqrt2}{4}.$$

Similarly, the cosine difference formula is

$$\cos(A-B)=\cos A\cos B+\sin A\sin B.$$

Thus

$$\cos15^{\circ}= \cos(45^{\circ}-30^{\circ}) = \cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}.$$

Substituting the same exact values, we find

$$\cos15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4} =\dfrac{\sqrt6+\sqrt2}{4}.$$

With these, return to the half-angle formula:

$$\cot7.5^{\circ}= \dfrac{1+\cos15^{\circ}}{\sin15^{\circ}} = \dfrac{1+\dfrac{\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}}.$$

Combine terms in the numerator:

$$1+\dfrac{\sqrt6+\sqrt2}{4}=\dfrac{4}{4}+\dfrac{\sqrt6+\sqrt2}{4} =\dfrac{4+\sqrt6+\sqrt2}{4}.$$

Therefore

$$\cot7.5^{\circ}= \dfrac{\dfrac{4+\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}} =\dfrac{4+\sqrt6+\sqrt2}{\sqrt6-\sqrt2}.$$

To eliminate the denominator’s square roots we multiply numerator and denominator by the conjugate $$(\sqrt6+\sqrt2):$$

$$\cot7.5^{\circ}= \dfrac{(4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2)} {(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)}.$$

The denominator simplifies first:

$$(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)=(\sqrt6)^2-(\sqrt2)^2=6-2=4.$$

Now expand the numerator term by term:

$$\begin{aligned} (4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2) &=4(\sqrt6+\sqrt2)+\sqrt6(\sqrt6+\sqrt2)+\sqrt2(\sqrt6+\sqrt2)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl((\sqrt6)^2+\sqrt6\sqrt2\bigr) +\bigl(\sqrt2\sqrt6+(\sqrt2)^2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl(6+\sqrt{12}\bigr) +\bigl(\sqrt{12}+2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +6+2+\sqrt{12}+\sqrt{12}\\[4pt] &=4\sqrt6+4\sqrt2 +8+2\sqrt{12}. \end{aligned}$$

Because $$\sqrt{12}=2\sqrt3,$$ we rewrite:

$$4\sqrt6+4\sqrt2+8+2\sqrt{12}=4\sqrt6+4\sqrt2+8+4\sqrt3.$$

Thus the numerator is $$8+4\sqrt2+4\sqrt3+4\sqrt6.$$ Dividing this by the denominator $$4$$ gives

$$\cot7.5^{\circ}=\dfrac{8+4\sqrt2+4\sqrt3+4\sqrt6}{4} =\;2+\sqrt2+\sqrt3+\sqrt6.$$

But $$\cot7.5^{\circ}=\cot\dfrac{\pi}{24},$$ so

$$\cot\dfrac{\pi}{24}=2+\sqrt2+\sqrt3+\sqrt6.$$

This matches option B in the list provided. Hence, the correct answer is Option 2.