Given that \( a = 30^{\circ} \) and \( b = 45^{\circ} \), calculate \( \tan(a + b) \) using the sum formula.

To find \( \tan(a + b) \) using the sum formula, we start with the angle sum formula for tangent: \[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] Now we need to calculate \( \tan a \) and \( \tan b \). 1. For \( a = 30^{\circ} \): \[ \tan 30^{\circ} = \frac{1}{\sqrt{3}} \approx 0.577 \] 2. For \( b = 45^{\circ} \): \[ \tan 45^{\circ} = 1 \] Now we can substitute these values into the formula: \[ \tan(30^{\circ} + 45^{\circ}) = \frac{\tan 30^{\circ} + \tan 45^{\circ}}{1 - \tan 30^{\circ} \tan 45^{\circ}} = \frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}} \cdot 1} \] Now, we simplify both the numerator and the denominator. **Numerator:** \[ \frac{1}{\sqrt{3}} + 1 = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{\sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{3}} \] **Denominator:** \[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \] Now, substituting back into the tangent formula: \[ \tan(30^{\circ} + 45^{\circ}) = \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \] To simplify further, we can rationalize the denominator by multiplying the numerator and denominator by \( \sqrt{3} + 1 \): \[ \tan(30^{\circ} + 45^{\circ}) = \frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \] **Denominator:** \[ (\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2 \] **Numerator:** \[ (1 + \sqrt{3})(\sqrt{3} + 1) = 1 \cdot \sqrt{3} + 1 \cdot 1 + \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1 = \sqrt{3} + 1 + 3 + \sqrt{3} = 4 + 2\sqrt{3} \] Thus, we have: \[ \tan(30^{\circ} + 45^{\circ}) = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \] Therefore, the value of \( \tan(30^{\circ} + 45^{\circ}) \) is \[ \boxed{2 + \sqrt{3}} \]