1. **State the problem:** Find the exact value of $\tan 165^\circ$ using a sum or difference formula. 2. **Recall the formula:** The tangent of a sum or difference of angles is given by $$\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$$ 3. **Choose angles:** Note that $165^\circ = 180^\circ - 15^\circ$, so we can use the difference formula: $$\tan 165^\circ = \tan(180^\circ - 15^\circ)$$ 4. **Use tangent difference formula:** $$\tan(180^\circ - \theta) = -\tan \theta$$ So, $$\tan 165^\circ = -\tan 15^\circ$$ 5. **Find $\tan 15^\circ$ using sum/difference:** $$15^\circ = 45^\circ - 30^\circ$$ Apply the difference formula: $$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$$ 6. **Recall exact values:** $$\tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}$$ 7. **Substitute values:** $$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$$ 8. **Simplify numerator and denominator:** Multiply numerator and denominator by $\sqrt{3}$ to rationalize: $$\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$ 9. **Rationalize denominator:** Multiply numerator and denominator by the conjugate $\sqrt{3} - 1$: $$\tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3})^2 - 2\sqrt{3} + 1}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$ 10. **Find $\tan 165^\circ$:** $$\tan 165^\circ = -\tan 15^\circ = -(2 - \sqrt{3}) = \sqrt{3} - 2$$ **Final answer:** $$\boxed{\tan 165^\circ = \sqrt{3} - 2}$$