Finding $\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right) $
assume $\displaystyle \frac{\pi}{14} = \theta$ then $\tan (14\theta) = 0$
wan,t be able to go further, some help me
We have $\tan \frac {\pi}{14} = \cot (\frac {\pi}{2}-\frac {\pi}{14}) = \cot \frac {3\pi}{7} $. Similarly, we have, $\tan \frac {3\pi}{14} = \cot \frac {2\pi}{7} $ and $\tan \frac {5\pi}{14} = \cot \frac {\pi}{7} $.
Now, our expression simplifies to evaluating $$P = \tan \frac {\pi}{14} \tan \frac {3\pi}{14} \tan \frac {5\pi}{14}$$ $$ = \cot \frac {\pi}{7}\cot \frac {2\pi}{7}\cot \frac {3\pi}{7} $$ $$= \frac {1}{\tan \frac {\pi}{7}\tan \frac {2\pi}{7}\tan \frac {3\pi}{7}} \tag {1}$$
If we let $\theta = \pi/7$, then we get, $7\theta = 4\theta + 3\theta = \pi \Rightarrow 4\theta = \pi -3\theta $. Thus, $\tan 4\theta = \tan (\pi -3\theta) = -\tan 3\theta $. So, we get, $$\frac {4\tan \theta - 4\tan^3 \theta}{1-6\tan^2 \theta + \tan^4 \theta} = -\frac {3\tan \theta -\tan^3 \theta}{1-3\tan^2 \theta} $$
To proceed, note that you will get a sixth degree equation whose roots are $\tan \frac {k\pi}{7} , k\in \{1,2,\cdots,6\} $. Also, notice that $\tan^2 \frac {k\pi}{7} = \tan^2 \frac {(7-k)\pi}{7}, k \in \{1,2,3\} $. Keep $x^2=y $, and $P $ will turn out to be reciprocal of the square root of the product of the roots in the new equation in terms of $y $ (why?). The answer is thus $\boxed {\displaystyle \frac {1}{\sqrt {7}}} $.
Hope it helps.