<s>[INST] Show that $\tan{(\pi/7)} \tan{(2\pi/7)}\tan{(3\pi/7)}=\sqrt{7}$ I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do
 [/INST] If $\theta = \frac{k\pi}{7}$ where $k = 1, 2, 3$, then $7\theta = k\pi$ and hence $4\theta = k\pi - 3\theta$. Thus $\tan(4\theta) = -\tan(3\theta)$. Expanding, and writing $t = \tan\theta$, we get 
$$ \frac{4t-4t^3}{1-4t^2 + t^4} = -\frac{3t-t^3}{1-3t^2} $$
Simplifying we get 
$$t^6 - 7t^4 + \cdots -7 = 0$$
The roots are $\tan \pi/7, \tan 2\pi/7, \ldots, \tan 6\pi/7$. Noting that $\tan 2\pi/7 = - \tan 5\pi/7, \tan 4\pi/7 = -\tan 3\pi/7$ etc, the roots are $\pm \tan \pi/7, \pm \tan 2\pi/7, \pm \tan 3\pi/7$. Thus the product of the roots is 
$$-\tan^2\pi/7 \tan^2 2\pi/7 \tan^2 3\pi/7 = -7$$ and since $\tan k\pi/7$ for $k = 1, 2, 3$ are all positive, we get 
$$ \tan \pi/7 \tan 2\pi/7 \tan 3\pi/7 = \sqrt{7}$$
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