MHB Proof: $\tan\,6^{\circ}\tan\,42^{\circ}\tan\,66^{\circ}\tan\78^{\circ}=1$

[kaliprasad]

prove that $\tan\,6^{\circ}\tan\,42^{\circ}\tan\,66^{\circ}\tan\,78^{\circ} = 1$

 

[anemone]

My solution:

Spoiler

We know that

$\sin 6^{\circ}\cdot \sin 54^{\circ}\cdot \sin 66^{\circ}=\dfrac{\sin 18^{\circ}}{4} $ and $\cos 6^{\circ}\cdot \cos 54^{\circ}\cdot \cos 66^{\circ}=\dfrac{\cos 18^{\circ}}{4} $

Dividing first by the second we get:

$\tan 6^{\circ}\cdot \tan 54^{\circ}\cdot \tan 66^{\circ}=\tan 18^{\circ} $

$\tan 6^{\circ}\cdot \tan 66^{\circ}=\dfrac{\tan 18^{\circ}}{\tan 54^{\circ}}$---(*)

Also, we have that

$\sin 18^{\circ}\cdot \sin 42^{\circ}\cdot \sin 78^{\circ}=\dfrac{\sin 54^{\circ}}{4} $ and $\cos 18^{\circ}\cdot \cos 42^{\circ}\cdot \cos 78^{\circ}=\dfrac{\cos 54^{\circ}}{4} $

Divide again these two equations yields $\tan 18^{\circ}\cdot \tan 42^{\circ}\cdot \tan 78^{\circ}=\tan 54^{\circ} $ and rearrange it to obtain $\dfrac{\tan 18^{\circ}}{\tan 54^{\circ}}=\dfrac{1}{\tan 42^{\circ}\cdot \tan 78^{\circ}}$ and substitute this into (*) the result follows and we're done.

 

[kaliprasad]

My solution:

Spoiler

We know that

$\sin 6^{\circ}\cdot \sin 54^{\circ}\cdot \sin 66^{\circ}=\dfrac{\sin 18^{\circ}}{4} $ and $\cos 6^{\circ}\cdot \cos 54^{\circ}\cdot \cos 66^{\circ}=\dfrac{\cos 18^{\circ}}{4} $

Dividing first by the second we get:

$\tan 6^{\circ}\cdot \tan 54^{\circ}\cdot \tan 66^{\circ}=\tan 18^{\circ} $

$\tan 6^{\circ}\cdot \tan 66^{\circ}=\dfrac{\tan 18^{\circ}}{\tan 54^{\circ}}$---(*)

Also, we have that

$\sin 18^{\circ}\cdot \sin 42^{\circ}\cdot \sin 78^{\circ}=\dfrac{\sin 54^{\circ}}{4} $ and $\cos 18^{\circ}\cdot \cos 42^{\circ}\cdot \cos 78^{\circ}=\dfrac{\cos 54^{\circ}}{4} $

Divide again these two equations yields $\tan 18^{\circ}\cdot \tan 42^{\circ}\cdot \tan 78^{\circ}=\tan 54^{\circ} $ and rearrange it to obtain $\dfrac{\tan 18^{\circ}}{\tan 54^{\circ}}=\dfrac{1}{\tan 42^{\circ}\cdot \tan 78^{\circ}}$ and substitute this into (*) the result follows and we're done.

good solution
here is mine

Spoiler

using $\tan\theta\tan(60^\circ -\theta)\tan(60^\circ + \theta) = \tan3\theta$
for a proof see below

we have taking $\theta=6^\circ$
$\tan6^\circ \tan54^\circ \tan66^\circ = \tan18^\circ\cdots{1}$
taking $\theta=18^\circ$
$\tan18^\circ \tan42^\circ \tan78^\circ = \tan54^\circ\cdots{2}$

multiplying above (1) and (2) we get

$\tan6^\circ \tan42^\circ \tan66^\circ \tan78^\circ=1 $

Spoiler

to prove

$\tan\theta\tan(60^\circ -\theta)\tan(60^\circ + \theta) = \tan3\theta$

$\tan(60^\circ -\theta)\tan(60^\circ + \theta)$
= $\dfrac{\tan60^\circ-\tan\theta}{1+ \tan60^\circ\tan \theta}\dfrac{\tan60^\circ+\tan\theta}{1- \tan60^\circ\tan \theta}$

= $\dfrac{\tan^260^\circ-\tan^2\theta}{1- \tan^260^\circ\tan^2 \theta}$
= $\dfrac{3-\tan^2\theta}{1- 3\tan^2 \theta}$so

$\tan\theta\tan(60^\circ -\theta)\tan(60^\circ + \theta)$

= $\dfrac{3\tan\theta-\tan^3\theta}{1- 3\tan^2 \theta}$
= $\tan3\theta$