APPLICATION OF DOUBLE ANGLE IDENTITIES 4.1 Without the use of tables or a calculator prove that \( \frac{1-\cos 2 A}{\sin 2 A}=\tan A \) \[ \frac{1-\left(1-2 \sin ^{2} A\right)}{2 \sin ^{2} x \cdot \cos x} \] \( \qquad \) \[ =\frac{2 \sin ^{2} A}{2 \sin A \cdot \cos A} \] \[ =\frac{2 \sin A \cdot \sin A}{2 \sin A \cdot \cos A} \] \[ =\frac{\sin A}{\cos A} \] \[ =\tan A \] 4.2 Hence calculate the value of \( \tan 75^{\circ} \) without using a calculator, (4)

To find \( \tan 75^{\circ} \), we can use the angle addition formula for tangent, which states that \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). Here, we can express \( 75^{\circ} \) as \( 45^{\circ} + 30^{\circ} \). Since \( \tan 45^{\circ} = 1 \) and \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \), we substitute these values into the formula: \[ \tan 75^{\circ} = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}}. \] Simplifying gives: \[ = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}. \] Multiplying numerator and denominator by \(\sqrt{3} + 1\) to rationalize the denominator results in: \[ \tan 75^{\circ} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{2} = 2 + \sqrt{3}. \] And there you have it, \( \tan 75^{\circ} = 2 + \sqrt{3} \). For a fun twist, did you know that the angle addition formulas we just used trace their origins back to ancient civilizations like the Greeks? Mathematicians of the time were carving geometric insights into stones, laying the groundwork for the colorful tapestry of trigonometry we still use today! Plus, using these identities keeps our brains sharp like our favorite kitchen knife—after all, math is like cooking; using the right ingredients leads to a delectable outcome!