Find the value of tan(π/7)·tan(2π/7)·tan(3π/7).

Step 1: Define the angle and use trigonometric identities

Let's define $\theta = \frac{\pi}{7}$ to simplify the expression. The problem then becomes finding the value of $\tan(\theta)\tan(2\theta)\tan(3\theta)$. We know that $7\theta = \pi$, which implies $4\theta = \pi - 3\theta$.

Step 2: Apply tangent to both sides

Taking the tangent of both sides of the equation $4\theta = \pi - 3\theta$, we get $\tan(4\theta) = \tan(\pi - 3\theta)$. Since $\tan(\pi - x) = -\tan(x)$, this simplifies to $\tan(4\theta) = -\tan(3\theta)$.

Step 3: Expand using tangent multiple angle formulas

We use the multiple angle formulas for tangent: $\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$, $\tan(3x) = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$, and $\tan(4x) = \frac{2\tan(2x)}{1 - \tan^2(2x)}$. Substituting these into the equation $\tan(4\theta) = -\tan(3\theta)$ gives the expanded form.

Step 4: Simplify the equation

Let $t = \tan\theta$. After cross-multiplication and simplification, we get a polynomial equation in terms of $t$. We can factor out $t$ from the left side of the equation $\tan(4\theta) = -\tan(3\theta)$ if $\tan\theta \neq 0$. Since $\theta = \pi/7$, $\tan\theta \neq 0$. The simplified equation is $\tan^6\theta - 21\tan^4\theta + 35\tan^2\theta - 7 = 0$.

Step 5: Relate to roots of a polynomial

Let $x = \tan^2\theta$. The equation becomes $x^3 - 21x^2 + 35x - 7 = 0$. The roots of this cubic equation are $\tan^2(\pi/7)$, $\tan^2(2\pi/7)$, and $\tan^2(3\pi/7)$.

Step 6: Apply Vieta's formulas

According to Vieta's formulas, for a cubic equation $ax^3 + bx^2 + cx + d = 0$, the product of the roots is $-d/a$. In our case, the product of the roots $\tan^2(\pi/7)$, $\tan^2(2\pi/7)$, and $\tan^2(3\pi/7)$ is $-(-7)/1 = 7$.

Step 7: Calculate the final product

Since $\pi/7$, $2\pi/7$, and $3\pi/7$ are all in the first quadrant, their tangent values are positive. Therefore, the product $\tan(\pi/7) \cdot \tan(2\pi/7) \cdot \tan(3\pi/7)$ must be positive. Taking the square root of the product of the squares of the tangents, we get $\sqrt{7}$.

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