TITLE: A trigonometric equation
QUESTION [8 upvotes]: Some years ago, I came across the following question:

Find the value of $\tan5°\tan55°\tan65°\tan75°$.

The value is 1 and I generalized the question as follows.

$\tan a°\tan b°\tan c°\tan d° =1 $ if $(a, b, c, d, )=(3p, 30+p, 90-p, 30-p)$

Here comes my question:

Are there $a, b, c, d \in \mathbb{N}$ such that $\tan a°\tan b°\tan c°\tan d° =2 $?

If yes, 

What are the values of $n \in \mathbb{N}$ so that there are integer solutions for    $\tan a°\tan b°\tan c°\tan d° =n $?

REPLY [4 votes]: Let  $\displaystyle \tan5x=\tan5A$
then $5x=r180^\circ+5A$ where $r$ is any inetger
$\displaystyle\implies x=r36^\circ+A$ where $r\equiv0,1,2,3,4\pmod 5\ \ \ \ (1)$
Again using this, $\displaystyle \tan5x=\frac{\tan^5x-10\tan^3x+5\tan x}{5\tan^4x-10\tan^2x+1}\  \   \   \ (2)$
$\displaystyle\implies  \frac{\tan^5x-10\tan^3x+5\tan x}{5\tan^4x-10\tan^2x+1}=\tan5A $
$\displaystyle\implies \tan^5x-5\tan5A\tan^4x-10\tan^3x+10\tan5A\tan^2x+5\tan x-\tan5A=0$ which is a five degree equation in $\tan x$
Using Vieta's formulas, $\displaystyle \prod_{0\le r\le4}\tan\left(r36^\circ+A\right)=\tan5A\ \ \ \ (3)$
$\displaystyle\implies\prod_{0\le r\le4,r\ne m}\tan\left(r36^\circ+A\right)=\frac{\tan5A}{\tan\left(m36^\circ+A\right)}$ where $m$ should be so chosen that $\tan\left(m36^\circ+A\right)\ne0$
Now,  $\displaystyle\prod_{0\le r\le4,r\ne m}\tan\left(r36^\circ+A\right)=n\ \ \  \  (4)$ if $\displaystyle\frac{\tan5A}{\tan\left(m36^\circ+A\right)}=n$
Let $\displaystyle m36^\circ+A=B\implies 5A=5(B-m36^\circ)=5B-m180^\circ\implies \tan5A=\tan5B$
$\displaystyle\implies \frac{\tan5B}{\tan B}=n$
$\displaystyle\implies n\tan B={\frac{\tan^5B-10\tan^3B+5\tan B}{5\tan^4B-10\tan^2B+1}}$
$\displaystyle\implies n={\frac{\tan^4B-10\tan^2B+5}{5\tan^4B-10\tan^2B+1}}$ as $\tan B\ne0$
On rearrangement we get, 
$\displaystyle (5n-1)\tan^4B-10(n-1)\tan^2B+n-5=0\ \  \  \  (5)$ which is clearly a Quadratic Equation in $\tan^2B$
As $B$ is real,  the discriminant must be $\ge0$(which is true for  all real $n$) and $\tan^2B\ge0$
So, any real value of $n$ that satisfies the above condition  can form relation $(4)$
Examples
$1:$ Let's start with $(3)$ and with known $A=3^\circ$
$\displaystyle\implies \tan3^\circ\tan39^\circ\tan75^\circ\tan111^\circ\tan147^\circ=\tan15^\circ\ \ \ \  (6A)$
Now, $\tan111^\circ=-\tan(180^\circ-111^\circ)=-\tan69^\circ$ and similarly, $\tan147^\circ=-\tan33^\circ$
$\displaystyle\implies \tan3^\circ\tan39^\circ\tan69^\circ\tan33^\circ=\tan15^\circ\cot75^\circ\ \ \ \  (6B)$
Now, $\cot75^\circ=\tan(90^\circ-75^\circ)=\tan15^\circ$ and $\tan15^\circ=\tan(60^\circ-45^\circ)=\cdots=2-\sqrt3$
Interestingly, we shall reach at $(6B)$ if we start with $A=39^\circ,69^\circ,33^\circ,75^\circ$
and we shall get the angles in complementary form if we start with $36^\circ m+15^\circ$ where $m$ is any integer
$2:$ Setting $A=6^\circ, \implies \tan6^\circ\tan42^\circ\tan78^\circ\tan114^\circ\tan150^\circ=\tan30^\circ\ \ \ \  (7A)$
Now, $\tan114^\circ=-\tan66^\circ$ and $\tan150^\circ=-\tan30^\circ$
$\displaystyle\implies  \tan6^\circ\tan42^\circ\tan78^\circ\tan66^\circ=1\ \ \ \  (7B)$
Again, we shall reach at $(7B)$ if we start with $A=42^\circ,78^\circ,66^\circ$
We shall get $1$ in the Right Hand Side, if $5A=u180^\circ\pm(r36^\circ+A)$ where $u,r$ are any integers
Try with any multiple of $3$ as $84^\circ,63^\circ$ etc. for $A$
Finally don't miss to try some values of $n$ in $(5)$