1. The problem is to find the value of $\tan 75^\circ$. 2. We use the tangent addition formula: $$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$ 3. Express $75^\circ$ as $45^\circ + 30^\circ$. 4. Substitute $a=45^\circ$ and $b=30^\circ$ into the formula: $$\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}$$ 5. Recall that $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$. 6. Substitute these values: $$\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$$ 7. Multiply numerator and denominator by $\sqrt{3}$ to rationalize: $$\tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$ 8. Multiply numerator and denominator by the conjugate of the denominator $\sqrt{3} + 1$: $$\tan 75^\circ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3})^2 + 2\sqrt{3} + 1}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2}$$ 9. Simplify the fraction: $$\tan 75^\circ = 2 + \sqrt{3}$$ Final answer: $$\boxed{2 + \sqrt{3}}$$