To find the exact value of $\cot 15^\circ$, we use the identity:

$\cot(15^\circ) = \frac{\cos(15^\circ)}{\sin(15^\circ)}$

We calculate $\cos 15^\circ$ and $\sin 15^\circ$ using the half-angle formulas:

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$

$= \left(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \times \frac{1}{2}\right)$

$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$

Similarly,

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

$= \left(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2} \times \frac{1}{2}\right)$

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Now,

$\cot 15^\circ = \frac{\cos 15^\circ}{\sin 15^\circ} = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{6} - \sqrt{2}}{4}}$

Canceling the denominator,

$\cot 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}$

Multiply by the conjugate of the denominator:

$\cot 15^\circ = \frac{(\sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}$

$= \frac{(\sqrt{6})^2 + 2\sqrt{6} \cdot \sqrt{2} + (\sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}$

$= \frac{6 + 2\sqrt{12} + 2}{6 - 2}$

$= \frac{8 + 4\sqrt{3}}{4}$

$= 2 + \sqrt{3}$

Thus, the exact value of $\cot 15^\circ$ is:

$\boxed{2 + \sqrt{3}}$

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Would you like a step-by-step breakdown of any part?

Here are some related problems for practice:

1. Find the exact value of $\tan 15^\circ$.
2. Prove that $\cot 75^\circ = 2 - \sqrt{3}$.
3. Derive the cotangent addition formula.
4. Find the exact value of $\cos 75^\circ$.
5. Solve $\cot x = 2 + \sqrt{3}$ for $x$ in degrees.

Tip: Rationalizing denominators often helps simplify trigonometric expressions!