Tan 75 Degrees 9Aceed

1. **State the problem:** We need to express $\tan 75^\circ$ in the form $a + b\sqrt{3}$ where $a$ and $b$ are integers. 2. **Recall the formula for tangent of sum of angles:** $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ 3. **Express $75^\circ$ as sum of $45^\circ$ and $30^\circ$:** $$\tan 75^\circ = \tan(45^\circ + 30^\circ)$$ 4. **Use known tangent values:** $$\tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}$$ 5. **Apply the formula:** $$\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$$ 6. **Simplify numerator and denominator by multiplying numerator and denominator by $\sqrt{3}$:** $$\tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$ 7. **Rationalize the denominator:** $$\tan 75^\circ = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{(\sqrt{3} + 1)^2}{2}$$ 8. **Expand numerator:** $$ (\sqrt{3} + 1)^2 = (\sqrt{3})^2 + 2 \times \sqrt{3} \times 1 + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} $$ 9. **Substitute back:** $$\tan 75^\circ = \frac{4 + 2\sqrt{3}}{2}$$ 10. **Simplify fraction:** $$\tan 75^\circ = \frac{\cancel{2} \times 2 + \cancel{2} \times \sqrt{3}}{\cancel{2}} = 2 + \sqrt{3}$$ 11. **Identify $a$ and $b$:** $$a = 2, \quad b = 1$$ **Final answer:** $\tan 75^\circ = 2 + 1 \times \sqrt{3}$ where $a=2$ and $b=1$.