Given that (tan 2A = frac{2 tan A}{1 – tan^{2} A}), evaluate (tan 15°), leaving your answer in surd form.
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} + 2sqrt{3} x – 1 = 0)
(x = frac{-(2sqrt{3}) pm sqrt{(2sqrt{3})^{2} – 4(1)(-1)}}{2(1)})
(x = frac{-2sqrt{3} pm sqrt{16}}{2} = frac{-2sqrt{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}).