4.2 Hence calculate the value of \( \tan 75^{\circ} \) without using a calculato
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To find \( \tan 75^{\circ} \), we can use the angle sum formula for tangent: \[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] Here, we can express \( 75^{\circ} \) as \( 45^{\circ} + 30^{\circ} \). So, let’s denote: \[ a = 45^{\circ} \quad \text{and} \quad b = 30^{\circ} \] We know that: \[ \tan 45^{\circ} = 1 \quad \text{and} \quad \tan 30^{\circ} = \frac{1}{\sqrt{3}} \] Now, substituting these values in: \[ \tan 75^{\circ} = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \] This simplifies to: \[ \tan 75^{\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \] To further simplify it, we can multiply the numerator and denominator by \( \sqrt{3} + 1 \): \[ \tan 75^{\circ} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \] So, the value of \( \tan 75^{\circ} \) is \( 2 + \sqrt{3} \).