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To find the exact value of \( \tan 75^{\circ} \), we can use the angle addition formula for tangent. We know that \( 75^{\circ} = 45^{\circ} + 30^{\circ} \). The tangent addition formula is given by: \[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] Setting \( a = 45^{\circ} \) and \( b = 30^{\circ} \): \[ \tan 45^{\circ} = 1 \quad \text{and} \quad \tan 30^{\circ} = \frac{1}{\sqrt{3}} \] Plugging these values into the formula: \[ \tan 75^{\circ} = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \] Now, simplifying the expression step by step: The numerator becomes: \[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \] The denominator becomes: \[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \] Putting it all together: \[ \tan 75^{\circ} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \] To simplify further, we can multiply the numerator and denominator by \( \sqrt{3} + 1 \): \[ \tan 75^{\circ} = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} \] Calculating each of these: The denominator becomes: \[ (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \] The numerator becomes: \[ (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \] Thus, we have: \[ \tan 75^{\circ} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \] Therefore, the exact value of \( \tan 75^{\circ} \) is: \[ \boxed{2 + \sqrt{3}} \]
