Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$ 
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ without using Limit as a sum.

$\bf{My\; Try::}$ Using the formula $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Replace $n\rightarrow 2n$
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
Now How can I calculate $$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)$$
and also How can I calculate $$\displaystyle \cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$$
Help required, Thanks
 A: I will also share my thoughts on your formula.
In moving from 
$$ \displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}} \ \ \ (1)$$
to
$$ \displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right) \ \ \ (2)$$
Notice the symmetry of $\sin $ function around $\pi/2$, we can see that for all $k=1,2,...,n$
$$ \displaystyle \sin\frac{(n-k)\pi}{2n} = \sin(\frac{(n-k)\pi}{2n}+\frac{\pi}{2})=\sin\frac{(n+k)\pi}{2n}$$
So we see that (2) is equal to the square root of (1) -- almost! (Almost because depending on $n$ being odd or even there may or may not be a perfect match-up of $k$ less that $n$ with those bigger than $n$.)
For $\cos$ 's use the identity $$ \sin (\frac{\pi}{2}-x)=\cos (x)$$ to see that for any $k=1,2,...,n$
$$ =\cos (\frac{k\pi}{2n})=\sin (\frac{\pi}{2}-\frac{k\pi}{2n})=\sin (\frac{(n-k)\pi}{2})$$
But $n-k$ will range over the same integers.
Notice then that in your expression first numerator then will be cancelled from last denominator. Then 2nd numerator with the second to last denominator... Then your left with $1$! and the answer to your limit is $1$ -- which was corroborated by the alternative calculation.
A: To complete the answer started by @Behnam, note that we can write
$$\begin{align}
\int_0^1 \log\left(\tan\left(\frac{\pi x}{2}\right)\right)\,dx&=\frac{2}{\pi}\int_0^{\pi/2}\log(\tan(x))\,dx \tag 1\\\\
&=\frac{2}{\pi}\int_0^{\pi/4}\log(\tan(x))\,dx+\frac{2}{\pi}\int_{\pi/4}^{\pi/2}\log(\tan(x))\,dx \tag 2\\\\
&=\frac{2}{\pi}\int_0^{\pi/4}\log(\tan(x))\,dx+\frac{2}{\pi}\int_0^{\pi/4}\log(\cot(x))\,dx \tag 3\\\\
&=0
\end{align}$$

NOTES:
In arriving at $(1)$, we enforced the substitution $\pi x/2\to x$.
In going from $(2)$ to $(3)$, we enforced the substitution $x\to \pi/2 -x$ along with using the identity $\tan(\pi/2-x)=\cot(x)$.

Therefore, we find that since the exponential function is continuous with $e^0=1$, the limit of interest is
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\prod_{k=1}^{n-1}\tan\left(\frac{k\pi}{2n}\right)\right)^{1/n}=1}$$
A: For fixed $n$, call the expression $A$, and take its $\log[.] $ to get:
$$ \log (A)=\frac{1}{n} \left(\log\tan \frac{\pi}{2n}+ \tan \frac{2\pi}{2n}+ .............. + \tan \frac{(n-1)\pi}{2n}\right) $$
$$ =\frac{1}{n} \sum_{k=1}^{n-1}\log\tan (\frac{\pi k}{2n})$$
Doesn't it look familiar?! Yes, Riemann sum as in the definition of definite integrals. Here, if $\ k/n$ is seen as our discretized parameter $x$, then
$$\lim_{n\rightarrow \infty} \log (A) = \int_0^1\log\tan(\frac{\pi }{2}x)dx$$
Once this value is calculated, $A$ will be the exponential of it. But how do we get $$ \int_0^1\log\tan(x)dx \ ?$$
Despite its nasty look, there seems to be hope to find an antiderivative. But I don't know right away.
A: First we have
\begin{align}
\lim_{n\to \infty}\frac1{n}\ln{\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)}
&=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\tan\frac{k\pi}{2n}\right)}
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\tan{x}}\:dx
\\
&=\frac{2}{\pi}\left(\int_0^{\pi/2}\ln\sin(x)dx -\int_0^{\pi/2}\ln\cos(x)dx\right)
\\
&=0
\end{align} 
The last step is through $y=\pi/2-x$.
Hence
$$
\lim_{n\to\infty}\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)^{\frac1{n}}=e^0=1
$$
