Without using calculator

We will start from, $cos(A + B) = cosA.cosB - sinA.sinB \Rightarrow tanA.tanB = 1 - \frac{cos(A +B)}{cosA.cosB}$ Using this property,

$tan(\frac{\pi}{7}) tan(\frac{2\pi}{7})+tan(\frac{2\pi}{7})tan(\frac{4\pi}{7}) +tan(\frac{4\pi}{7}) tan(\frac{\pi}{7})$

$= 3 - (\frac{cos\frac{3\pi}{7}}{cos\frac{\pi}{7}.cos\frac{2\pi}{7}} + \frac{cos(\frac{6\pi}{7})}{cos\frac{2\pi}{7}.cos\frac{4\pi}{7}} + \frac{cos(\frac{5\pi}{7})}{cos\frac{4\pi}{7}.cos\frac{\pi}{7}})$

$= 3 - \frac{cos(4\frac{\pi}{7})cos(\pi - 4\frac{\pi}{7}) + cos(2\frac{\pi}{7})cos(\pi - 2\frac{\pi}{7}) + cos(\frac{\pi}{7})cos(\pi - \frac{\pi}{7})}{cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}}$

$= 3 + \frac{cos^2(4\frac{\pi}{7}) + cos^2(2\frac{\pi}{7}) + cos^2(\frac{\pi}{7})}{cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}}$

Now, $cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7} = -\frac{1}{8}$ Proof given below.

$= 3 - 8(cos^2(4\frac{\pi}{7}) + cos^2(2\frac{\pi}{7}) + cos^2(\frac{\pi}{7}))$ $\because 2cos^2A = 1 + cos2A$

$= 3 - 4(3 + cos(\frac{2\pi}{7}) + cos(\frac{4\pi}{7}) + cos(\frac{8\pi}{7}))$

$= 3 - 4(3 + cos(\frac{2\pi}{7}) + cos(\frac{4\pi}{7}) + cos(\frac{6\pi}{7}))$

Now, $cos(\frac{2\pi}{7}) + cos(\frac{4\pi}{7}) + cos(\frac{6\pi}{7}) = -\frac{1}{2}$ Proof given below.

So, $tan(\frac{\pi}{7}) tan(\frac{2\pi}{7})+tan(\frac{2\pi}{7})tan(\frac{4\pi}{7}) +tan(\frac{4\pi}{7}) tan(\frac{\pi}{7})$

$= 3 - 4 \times (3 - \frac{1}{2})$

$= -7$

Proof 1 : $cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}$

Proof 2 :