<s>[INST] 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
 [/INST] 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.
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