1. The problem is to find the value of $\tan 15^\circ$. 2. We use the angle subtraction formula for tangent: $$\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$$ 3. We can express $15^\circ$ as $45^\circ - 30^\circ$. 4. Substitute $a = 45^\circ$ and $b = 30^\circ$ into the formula: $$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$$ 5. Recall the exact values: $$\tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}$$ 6. Substitute these values: $$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$$ 7. Multiply numerator and denominator by $\sqrt{3}$ to rationalize: $$\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$ 8. Multiply numerator and denominator by the conjugate of the denominator $\sqrt{3} - 1$: $$\tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3})^2 - 2\sqrt{3} + 1}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2}$$ 9. Simplify the fraction: $$\tan 15^\circ = 2 - \sqrt{3}$$ Final answer: $$\boxed{\tan 15^\circ = 2 - \sqrt{3}}$$