[Heir of Ramanujan]

10. เฉลยวิธีทำของผม
จะพิสูจน์ว่า $tan\,\theta\,tan\,(60^{\circ}-\theta)\,tan\,(60^{\circ}+\theta) = tan\,3\theta$
$\displaystyle{tan\,\theta\,tan\,(60^{\circ}-\theta)\,tan\,(60^{\circ}+\theta) = tan\,\theta\,\left[\frac{tan\,60^{\circ}-tan\,\theta}{1+tan\,60^{\circ}\,tan\,\theta}\right]\,\left[\frac{tan\,60^{\circ}+tan\,\theta}{1-tan\,60^{\circ}\,tan\,\theta}\right]}$
$\displaystyle{= tan\,\theta\,\left[\frac{\sqrt{3}-tan\,\theta}{1+\sqrt{3}\,tan\,\theta}\right]\,\left[\frac{\sqrt{3}+tan\,\theta}{1-\sqrt{3}\,tan\,\theta}\right]}$
$\displaystyle{= tan\,\theta\,\left[\frac{3-tan^2\,\theta}{1-3\,tan^2\,\theta}\right]}$
$\displaystyle{= \frac{3\,tan\,\theta-tan^3\,\theta}{1-3\,tan^2\,\theta} = tan\,3\theta}$

ดังนั้น
$\displaystyle{L.S. = tan\,50^{\circ}\,tan\,60^{\circ}\,tan\,70^{\circ} = \frac{tan\,10^{\circ}\,tan\,(60^{\circ}-10^{\circ})\,tan\,(60^{\circ}+10^{\circ})\,tan\,60^{\circ}}{tan\,10^{\circ}}}$
$\displaystyle{= \frac{tan\,30^{\circ}\,tan\,60^{\circ}}{tan\,10^{\circ}} = \frac{1}{tan\,10^{\circ}} = tan\,80^{\circ} = R.S.}$