Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving

Explanation

\(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\)

\(\tan 30 = \tan 2(15°)\)

\(\frac{\sqrt{3}}{3} = \frac{2 \tan 15}{1 - \tan^{2} 15°}\)

\(\frac{\sqrt{3}}{6} = \frac{\tan 15}{1 - tan^{2} 15°}\)

Let \(\tan 15° = x\)

\(\frac{\sqrt{3}}{6} = \frac{x}{1 - x^{2}}\)

\(\sqrt{3} (1 - x^{2}) = 6x \implies \sqrt{3} - \sqrt{3} x^{2} = 6x \)

\(\sqrt{3} x^{2} + 6x - \sqrt{3} = 0\)

Divide through by the coefficient of x\(^{2}\).

\(x^{2} + \frac{6}{\sqrt{3}} x - 1 = 0\)

\(\implies x^{2} + 2\sqrt{3} x - 1 = 0\)

\(x = \frac{-(2\sqrt{3}) \pm \sqrt{(2\sqrt{3})^{2} - 4(1)(-1)}}{2(1)}\)

\(x = \frac{-2\sqrt{3} \pm \sqrt{16}}{2} = \frac{-2\sqrt{3} \pm 4}{2}\)

\(x = -\sqrt{3} + 2 ; x = -\sqrt{3} - 2\)

Since 15° is in the first quadrant, \(\tan 15° > 0\)

\(\therefore \tan 15° = -\sqrt{3} + 2 \equiv 2 - \sqrt{3}\).