WAEC 2009 · Paper 2 · Q4

(a)
Given that $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ , evaluate $\tan 15^\circ$ , leaving your answer in surd form.

Worked solution (try it first)

Put $A = 15^\circ$ : $\tan 30^\circ = \dfrac{2\tan 15^\circ}{1 - \tan^2 15^\circ}$ .
Let $t = \tan 15^\circ$ and use $\tan 30^\circ = \dfrac{1}{\sqrt3}$ : $\dfrac{1}{\sqrt3} = \dfrac{2t}{1 - t^2}$ .
Cross-multiply: $1 - t^2 = 2\sqrt3\,t$ , so $t^2 + 2\sqrt3\,t - 1 = 0$ .
Use the formula: $t = \dfrac{-2\sqrt3 \pm \sqrt{12 + 4}}{2}$
$= \dfrac{-2\sqrt3 \pm 4}{2}$
$= -\sqrt3 \pm 2$ .
$15^\circ$ is acute, so $\tan 15^\circ$ is positive.
Take the $+$ sign.
So $\tan 15^\circ = 2 - \sqrt3$ .