Is Drawing A Regular Septagon Necessary?

$\begin{aligned} \tan \frac \pi 7 \tan \frac {2 \pi}7 \tan \frac {3 \pi}7 & = \frac {\sin \frac \pi 7 \sin \frac {2 \pi}7 \sin \frac {3 \pi}7}{\cos \frac \pi 7 \cos \frac {2 \pi}7 \cos \frac {3 \pi}7} = \frac AB \end{aligned}$

Using the identity:

$\begin{aligned} \prod_{k=1}^{n-1} \sin \left(\frac {k \pi}n \right) & = \frac n {2^{n-1}} \\ \implies \prod_{k=1}^6 \sin \left(\frac {k \pi}7 \right) & = \frac 7 {2^6} \\ \sin \frac \pi 7 \sin \frac {2 \pi}7 \sin \frac {3 \pi}7 \color{#3D99F6}{ \sin \frac {4\pi}7 \sin \frac {5 \pi}7 \sin \frac {6 \pi}7} & = \frac 7 {2^6} & \small \color{#3D99F6}{\text{As }\sin (\pi - x) = \sin x} \\ \sin \frac \pi 7 \sin \frac {2 \pi}7 \sin \frac {3 \pi}7 \color{#3D99F6}{\sin \frac {3\pi}7 \sin \frac {2 \pi}7 \sin \frac \pi 7} & = \frac 7 {2^6} \\ \implies A^2 & = \frac 7 {2^6} \\ A & = \frac {\sqrt 7}{2^3} \end{aligned}$

$\begin{aligned} B & = \cos \frac \pi 7 \cos \frac {2 \pi}7 \color{#3D99F6}{ \cos \frac {3 \pi}7} & \small \color{#3D99F6}{\text{As }\cos (\pi - x) = -\cos x} \\ & = \cos \frac \pi 7 \cos \frac {2 \pi}7 \color{#3D99F6}{\left(- \cos \frac {4 \pi}7\right)} \\ & = - \frac {{\color{#3D99F6}{\sin \frac \pi 7}} \cos \frac \pi 7 \cos \frac {2 \pi}7 \cos \frac {4 \pi}7}{\color{#3D99F6}{\sin \frac \pi 7}} \\ & = - \frac {\sin \frac {2 \pi}7 \cos \frac {2 \pi}7 \cos \frac {4 \pi}7}{2 \sin \frac \pi 7} \\ & = - \frac {\sin \frac {4 \pi}7 \cos \frac {4 \pi}7}{2^2 \sin \frac \pi 7} \\ & = - \frac {\sin \frac {8 \pi}7}{2^3 \sin \frac \pi 7} \\ & = - \frac {-\sin \frac \pi 7}{2^3 \sin \frac \pi 7} \\ & = \frac 1{2^3} \end{aligned}$

Therefore,

$\begin{aligned} \tan \frac \pi 7 \tan \frac {2 \pi}7 \tan \frac {3 \pi}7 & = \frac AB = \frac {\frac {\sqrt 7}{2^3}}{\frac 1{2^3}} = \sqrt 7 \end{aligned}$

$\implies A = \boxed{7}$ .

Since $\begin{array}{rcl} \sin7x & = & \displaystyle \mathfrak{Im}\big[(\cos x + i\sin x)^7\big] \; = \; 7\cos^6x \sin x - 35\cos^4x \sin^3x + 21\cos^2x\sin^5x - \sin^7x \\ & = & \sin^7x\big[7\cot^6x - 35\cot^4x + 21\cot^2x - 1\big] \; = \; \sin^7xF(\cot^2x) \end{array}$ where $F(X)$ is the polynomial $F(X) \; =\; 7X^3 - 35X^2 + 21X - 1$ we see that the roots of $F(X)$ are $\cot^2\tfrac{j\pi}{7}$ for $1 \le j \le 3$ , and hence $\cot^2\tfrac{\pi}{7} \cot^2\tfrac{2\pi}{7} \cot^2 \tfrac{3\pi}{7} = \tfrac17$ . Since all three angles are acute, it follows that $\tan\tfrac{\pi}{7} \tan\tfrac{2\pi}{7} \tan\tfrac{3\pi}{7} = \sqrt{7}$ , making the answer $\boxed{7}$ .