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 $?
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 $(5)$
This answers the case $n\neq 1$. Without considering $\tan60\tan60\tan x\tan(90-x)=3$, the code
#include <stdio.h>
#include <math.h>
int main ()
{
double result;
double tantable[90];
int a=0,b=0,c=0,d=0;
for(a=1;a<90;++a)
tantable[a]=tan(a*M_PI/180.0d);
for(a=89;a>0;--a)
for(b=a;b>0;--b)
for(c=b;c>0;--c)
for(d=c;d>0;--d){
result = tantable[a]*tantable[b]*tantable[c]*tantable[d];
int appresult=round(result);
if(a+d==90&&b==60&&c==60) continue;//tan(x)tan60tan60tan(90-x)
if(c+d==90&&a==60&&b==60) continue;//tan60tan60tan(x)tan(90-x) [just a permutation]
if(appresult>1 && (appresult-result<0.000001) && (result-appresult<0.000001))
printf("%d,%d,%d,%d->%lf\n",a,b,c,d,result);
}
return 0;
}
Produces the following log(commented by me):
89,36,34,14->7.000000//No. Error is beyond precision
88,78,37,31->60.999999//No
80,60,40,20->3.000000//Yes! (1)
79,74,44,22->7.000000//No. Error is beyond precision
78,59,57,53->16.000000//No. Error is beyond precision
76,76,64,23->14.000000//No. Error is beyond precision
73,73,61,25->9.000000//No. Error is beyond precision
72,72,36,36->5.000000//Yes!(2)
60,60,60,60->9.000000//Yes!(3)
Verification of the identities: $(1)$: $$\tan(30+x)\tan(3x)\tan(30-x)\tan(90-x)=1$$ $$\implies \tan40\tan30\tan20\tan80=1$$ $$\implies \tan40\tan20\tan80=\tan60$$ $$\implies (1)=\tan60\tan60=3$$
$(2)$: We know(or if skeptic, we can verify by setting $\sin3x=\cos2x$) $$\tan18=\sqrt{\frac{5-2\sqrt{5}}{5}}$$ By double angle formula, we get $$\implies\tan36=\sqrt{5-2\sqrt{5}}$$ And by $\frac{1}{\tan{18}}=\tan72$ $$\implies\tan72=\sqrt{\frac{5}{5-2\sqrt{5}}}$$ $$\implies\tan72\tan36=\sqrt{5}\implies\tan72\tan72\tan36\tan36=5$$ And $(3)$ is trivial.
So, we see that the only other such identities are of the forms: $$\tan(x)\tan(60)\tan(60)\tan(90-x)=3$$ $$\tan(80)\tan(60)\tan(40)\tan(20)=3$$ $$\tan(72)\tan(72)\tan(36)\tan(36)=5$$ $$\tan(60)\tan(60)\tan(60)\tan(60)=9$$
Progress in the $n=1$ case(does not consider known cases):
#include <stdio.h>
#include <math.h>
int main ()
{
double result;
double tantable[90];
int a=0,b=0,c=0,d=0;
for(a=1;a<90;++a)
tantable[a]=tan(a*M_PI/180.0);
for(a=89;a>0;--a)
for(b=a;b>0;--b)
for(c=b;c>0;--c)
for(d=c;d>0;--d){
result = tantable[a]*tantable[b]*tantable[c]*tantable[d];
if((1-result<0.000001) && (result-1<0.000001)&&(a+d!=90)&&(b+c!=90)){
if((d%3==0) && (b-d/3)==30 && (c+d/3)==30&&(a+d/3)==90) continue;
if(a+3*d==90 && b-d==60&&c+d==60) continue;
if(a-d==60 && b+d==60&&c+3*d==90) continue;
if(a-d==60 && c+d==60&&b+3*d==90) continue;
printf("%d,%d,%d,%d->%f\n",a,b,c,d,result);
}
}
return 0;
}
These are the remaining cases (equivalences commented by me, they are apparently true)
87,51,15,9->1.000000//1
87,27,21,15->1.000000 //2
84,48,24,12->1.000000//3
81,75,39,3->1.000000//1
81,57,21,15->1.000000//4
78,66,42,6->1.000000//3
75,69,63,3->1.000000//2
75,69,33,9->1.000000//4
75,39,33,27->1.000000//5
63,57,51,15->1.000000//5
We can reduce the number of statements to be proved:
Assuming $(1)$ $$\tan87\tan51\tan15\tan9=1\implies\tan51\tan15\tan9=\tan3 \tag{a.1}$$ $$\tan(30-x)\tan(30+x)\tan3x=\tan x\implies\tan27\tan33\tan9=\tan3 \tag{a.2}$$ $$(\text{a}.1),(\text{a}.2)\implies\tan51\tan15=\tan27\tan33\iff (5)$$
Assuming $(4)$ $$\tan75\tan69\tan33\tan9=\implies\tan75\tan69\tan33=\tan81 \tag{b.1}$$ $$\tan(60+x)\tan(60-x)\tan x=\tan3x\implies\tan87\tan33\tan27=\tan81 \tag{b.2}$$ $$(\text{b}.1),(\text{b}.2)\implies\tan75\tan69=\tan87\tan27\iff (2)$$
So, we just need to prove or disprove $$\tan87\tan51\tan15\tan9=1\tag{1}$$ $$\tan75\tan69\tan33\tan9=1\tag{2}$$ $$\tan78\tan66\tan42\tan6=1\tag{3}$$
Verification of $(3)$ $$\tan(60+x)\tan(60-x)\tan(x)=\tan3x\implies\tan78\tan42\tan18=\tan54$$ $$\implies\tan78\tan42=\tan54\tan72 \tag{3.1}$$ $$\tan(60+x)\tan(60-x)\tan(x)=\tan3x\implies\tan66\tan54\tan6=\tan18$$ $$\implies\tan72\tan6\tan66\tan54=1 \tag{3.2}$$ $$(3.1),(3.2)\implies\tan6\tan66\tan78\tan42\iff (3)$$ I have not been able to prove/disprove the other identities until now.
To answer, the first question, from this, I shall utilize $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$
Put $x=5^\circ$ to get $$\tan5^\circ\tan55^\circ\tan65^\circ=\tan15^\circ$$
Now, $\displaystyle\tan15^\circ=\cot(90^\circ-15^\circ)=\cot75^\circ=\frac1{\tan75^\circ}$
