Find the exact value of the expression. \[ \frac{\tan 15^{\circ}+\tan 30^{\circ}}{1-\tan 15^{\circ} \tan 30^{\circ}} \] \( \frac{\tan 15^{\circ}+\tan 30^{\circ}}{1-\tan 15^{\circ} \tan 30^{\circ}}=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To solve the expression \[ \frac{\tan 15^{\circ}+\tan 30^{\circ}}{1-\tan 15^{\circ} \tan 30^{\circ}}, \] we begin by recalling the tangent values for \(15^{\circ}\) and \(30^{\circ}\). 1. **Calculate \(\tan 30^{\circ}\)**: \[ \tan 30^{\circ} = \frac{1}{\sqrt{3}}. \] 2. **Calculate \(\tan 15^{\circ}\)** using the tangent subtraction formula: \[ \tan(15^{\circ}) = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}. \] Simplifying the numerator: \[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}. \] Simplifying the denominator: \[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}. \] Thus, \[ \tan 15^{\circ} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}. \] Now, substituting \(\tan 15^{\circ}\) and \(\tan 30^{\circ}\) into the original expression: \[ \frac{\frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{1}{\sqrt{3}}}{1 - \left(\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) \left(\frac{1}{\sqrt{3}}\right)}. \] 3. **Combine the terms in the numerator**: Finding a common denominator (\(\sqrt{3} + 1\)): \[ \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{1}{\sqrt{3}} = \frac{(\sqrt{3} - 1) \cdot \sqrt{3} + (\sqrt{3} + 1)}{\sqrt{3}(\sqrt{3} + 1)}. \] Simplifying the numerator: \[ \sqrt{3}(\sqrt{3} - 1) + (\sqrt{3} + 1) = 3 - \sqrt{3} + \sqrt{3} + 1 = 4. \] So, the numerator becomes: \[ \frac{4}{\sqrt{3}(\sqrt{3} + 1)}. \] 4. **Now for the denominator**: \[ 1 - \left(\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\cdot\frac{1}{\sqrt{3}}\right) = 1 - \frac{\sqrt{3} - 1}{3 + \sqrt{3}}. \] Simplifying: \[ 1 = \frac{3 + \sqrt{3}}{3 + \sqrt{3}} \implies \frac{3 + \sqrt{3} - (\sqrt{3} - 1)}{3 + \sqrt{3}} = \frac{3 + \sqrt{3} - \sqrt{3} + 1}{3 + \sqrt{3}} = \frac{4}{3 + \sqrt{3}}. \] 5. **Putting it all together**: Now we can simplify the entire expression: \[ \frac{\frac{4}{\sqrt{3}(\sqrt{3} + 1)}}{\frac{4}{3 + \sqrt{3}}} = \frac{4}{\sqrt{3}(\sqrt{3} + 1)} \cdot \frac{3 + \sqrt{3}}{4} = \frac{3 + \sqrt{3}}{\sqrt