Hence calculate the value of \( \tan 75^{\circ} \) without using a calcul
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To calculate \( \tan 75^{\circ} \), we can utilize the angle sum identity. Notably, \( 75^{\circ} \) can be expressed as \( 45^{\circ} + 30^{\circ} \). According to the formula for \( \tan(a + b) \): \[ \tan(75^{\circ}) = \tan(45^{\circ} + 30^{\circ}) = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} \] Given that \( \tan 45^{\circ} = 1 \) and \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \), we can substitute these values: \[ \tan(75^{\circ}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \] Now, simplify the fractions. Both the numerator and denominator can be multiplied by \( \sqrt{3} \): \[ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \] To simplify further, multiply the numerator and denominator by the conjugate of the denominator: \[ \tan(75^{\circ}) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \] Therefore, the value of \( \tan 75^{\circ} \) is \( 2 + \sqrt{3} \).