How can i solve this? the exercise is:
$$\tan{20^{\circ}}\cdot\tan{40^{\circ}}\cdot\tan80^{\circ}$$
I have to define the exact answer. I have no idea how to start this exercise so please help. I think that would be so easy if I just multiply them. Thank for your help.
$\tan20 \times \tan40 \times \tan80$
=$\frac{\sin40 \sin80 \sin20}{\cos40 \cos80 \cos20}$
Multiply numerator and denominator by 2,
=$\frac{(2\sin80 \sin40) \sin20}{(2\cos80 \cos40)\cos20}$
=$\frac{(\cos40- \cos120) \sin20}{(\cos120 + \cos40) \cos20}$
=$\frac{(2\cos40 + 1)\sin20}{(2\cos40 - 1) \cos20}$
=$\frac{(2\cos40\sin20) + \sin20}{(2\cos40\cos20) - \cos20}$
=$\frac{\sin60 - \sin20 + \sin20}{\cos60 + \cos20 - \cos20}$
=$\frac{\sin60}{\cos60}$
= $\tan 60$
Formulas-
2sinA sinB = cos(A-B) - cos(A+B)
2cosA cosB = cos(A+B) + cos(A-B)
2cosA sinB = sin(A+B) - sin(A-B)
Hint:-
$$\sin\theta\cdot\sin(60^{\circ}-\theta)\cdot\sin(60^{\circ}+\theta)=\dfrac{1}{4}(3\sin\theta-4\sin^3\theta)=\dfrac{1}{4}\sin{3\theta}\tag{1}$$
$$\cos\theta\cdot\cos(60^{\circ}-\theta)\cdot\cos(60^{\circ}+\theta)=\dfrac{1}{4}(4\cos^3\theta-3\cos\theta)=\dfrac{1}{4}\cos{3\theta}\tag{2}$$
$$\dfrac{(1)}{(2)}\implies \tan\theta\cdot\tan(60^{\circ}-\theta)\cdot\tan(60^{\circ}+\theta)=\tan{3\theta}$$
Now find that value of $\theta$ which gives you $\tan{20^{\circ}}\cdot\tan{40^{\circ}}\cdot\tan80^{\circ}$ in LHS