Post this for two years later

Note first that for any odd positive integer $n$ we have that

$\tan\left(\dfrac{k\pi}{n}\right) = -\tan\left(\dfrac{(n - k)\pi}{n}\right)$ for $1 \le k \le \dfrac{n - 1}{2}.$

So in the product $\displaystyle\prod_{k=1}^{n-1} \tan\left(\dfrac{k\pi}{n}\right)$ we can combine $\dfrac{n - 1}{2}$ pairs of terms to see that this product is equal to

$\displaystyle\prod_{k=1}^{\frac{n - 1}{2}} (-1)^{\frac{n - 1}{2}} \tan^{2}\left(\dfrac{k\pi}{n}\right) = (-1)^{\frac{n-1}{2}}\left(\prod_{k=1}^{\frac{n-1}{2}} \tan\left(\dfrac{k\pi}{n}\right)\right)^{2}$ for odd $n.$

But according to formula (35) here this is just equal to $(-1)^{\frac{n-1}{2}}(\sqrt{n})^{2} = (-1)^{\frac{n-1}{2}}n.$

So in this case the answer is $(-1)^{1008}*2017 = \boxed{2017}.$

In anticipation of the Challenge Master's response, I will make an attempt at proving formula (35) shortly .....

Edit: Formulas and their respective proofs for similar products involving the sine and cosine functions are provided here . Taken together, these provide us with the identity in formula (35). Perhaps there may be a more direct proof for the tan product that doesn't rely on finding the sine and cosine products first.

I think an easier way out of this problem will be to use the identity $\prod _{ r=1 }^{ n-1 }{ \tan { \frac { r\pi }{ n } } } =\frac { n }{ \sin { \frac { n\pi }{ 2 } } }$ which we can easily derive.

one can do this by complex numbers by using equation x^n-1=0 where n is 2017 and roots are e^(2kpi/n). k belongs to {0,1,....,2016}

My solution is quite long but maybe it'll be helpful.

I consider separately the product of sine and of cosine.

Consider first

$P=\prod_{k=1}^{n-1}\sin{\frac{\pi k}{n}}=\prod_{k=1}^{n-1}\sin{\frac{2\pi k}{2n}}$ .

Note that

$\sin{\frac{2\pi k}{2n}}=-\sin{\frac{2\pi(2n-k)}{2n}}$

and, excluding the case $k=n$ for which the sine is equal to $\sin{\pi}=0$ , I can then write:

$\prod_{{k=1}{k\neq n}}^{2n-1} \sin{\frac{2k\pi}{2n}}=(-1)^{n-1}P^2$

Now consider the $(2n)^{th}$ roots of unity:

$\omega_{k}=\omega^{k}=e^{\frac{2\pi ik}{2n}}$ for $1\leq k \leq 2n-1$ and $k\neq n$ ,

we of course have $\sin{\frac{2k\pi}{2n}} = \frac{\omega^{k}-\omega^{-k}}{2i}$ and so:

$\prod_{{k=1}{k\neq n}}^{2n-1} \sin{\frac{2k\pi}{2n}}=\frac{1}{2^{2(n-1)}i^{2(n-1)}}\prod_{{k=1}{k\neq n}}^{2n-1}(\omega^{k}-\omega^{-k})$

Remind that the $(2n)^{th}$ roots of unity are the solution of the equation $f(x)=\sum_{s=0}^{2n-1}x^s =0$ but we are excluding the $k=n$ root which is $e^{i\pi}=-1$ . So the roots that we are considering are the $2n-2$ roots of the equation $g(x)=f(x)/(x+1)=\sum_{m=0}^{n-1}x^{2m} =0$ . Now from Vieta's we have that $\prod_{{k=1}{k\neq n}}^{2n-1}\omega_{k}=1$ and so we can rewrite our product as:

$\prod_{{k=1}{k\neq n}}^{2n-1}(\omega^{2k}-1)= \prod_{{k=1}{k\neq n}}^{2n-1}(\omega_k-1)(\omega_k+1) = \prod_{{k=1}{k\neq n}}^{2n-1}(1-\omega_k) \prod_{{k=1}{k\neq n}}^{2n-1}((-1)-\omega_k)$ .

Now finally we have that

$\prod_{{k=1}{k\neq n}}^{2n-1}(1-\omega_k)=g(1)$

and

$\prod_{{k=1}{k\neq n}}^{2n-1}(-1-\omega_k)=g(-1)$ .

But $g(1)=g(-1)=n$ and we can write:

$\prod_{{k=1}{k\neq n}}^{2n-1} \sin{\frac{2k\pi}{2n}}=\frac{(-1)^{n-1}}{2^{2(n-1)}}n^2=(-1)^{n-1}P^2$ and so

$P=\frac{n}{2^{n-1}}$

I think it is possible to proceed in a similar way also for the product of cosine, but I want to use Chebyshev polynomials of the second kind which are defined as

$U_{n}(cos{\theta})=\frac{\sin{(n+1)\theta}}{\sin{\theta}}$ .

From the definition t is simple to prove that the zeros of $U_{n}(x)$ are

$x_{k}=\cos{\frac{k}{n-1}\pi}$ for $k=1,...,n$

and also it is simple to prove the reoccurrence relation

$U_{n}(x)=xU_{n-1}(x)+T_{n}(x)$

where $T_{n}$ are the Chebyshev polynomials of the first kind.

This relation can be used to prove that the constant term of $U_{n}(x)$ is $0$ if $n$ is odd, $1$ if $n=4m$ and $-1$ if $n=4m+2$ and that the leading term is $2^n$ .

Again, using Vieta's we see that $\prod_{k=1}^{n} \cos{\frac{k}{n-1}\pi}$ are equal to the constant term of $U_{n}(x)$ over the leading term. In our case we have:

$\prod_{k=1}^{n-1} \cos{\frac{k}{n}\pi} = \frac{U_{n-1}(0)}{2^{n-1}} =\frac{1}{2^{n-1}}$ .

In the last step I used the fact that for us $n=2017$ and so $n-1 = 4*502$ .

Putting all together:

$\prod_{k=1}^{n-1}\sin{\frac{\pi k}{n}} \frac{1}{\prod_{k=1}^{n-1}cos{\frac{\pi k}{n}}} = \frac{n}{2^{n-1}}2^{n-1}=n=2017$ .

Sir your problems are too awesome