[กิตติ]

ข้อ 2....ทำได้ 2แบบ
แบบที่1
$\tan 20^\circ \tan 40^\circ \tan 80^\circ$

$=\frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} $

$=\frac{\sin 80^\circ (\cos 20^\circ - \cos 60^\circ)}{\cos 80^\circ (\cos 20^\circ + \cos 60^\circ)} $

$=\frac{2\sin 80^\circ \cos 20^\circ-2\sin 80^\circ \cos 60^\circ}{2\cos 80^\circ \cos 20^\circ+2\cos 80^\circ \cos 60^\circ} $

$=\frac{\sin 100^\circ+\sin 60^\circ-\sin 80^\circ}{\cos 100^\circ+\cos 60^\circ+\cos 80^\circ} $

$\sin 100^\circ =\sin80^\circ,\cos 100^\circ= -\cos 80^\circ$

$=\frac{\sin 60^\circ}{\cos60^\circ} $

$=\tan 60^\circ = \sqrt{3} $

แบบที่ 2....มันมีสูตรว่า
$\tan (60-x)^\circ \tan x^\circ \tan (60+x)^\circ= \tan 3x^\circ $

แทน $x=20^\circ$

$\tan (60-20)^\circ \tan20^\circ \tan (60+20)^\circ= \tan (3\times 20)^\circ$

$\tan 40^\circ \tan 20^\circ \tan 80^\circ = \tan 60^\circ = \sqrt{3} $