Question

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

• A. 7
• B. -7
• C. 3
• D. -3

Hint:

For sums of tangent products involving angles $\frac{k\pi}{n}$, use properties of roots of unity or trigonometric identities. The sum of pairwise tangent products often equals $-n$ for specific angles.

Answer 1

The Correct Option is B

To solve the problem $\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{4\pi}{7}\right) + \tan\left(\frac{4\pi}{7}\right)\tan\left(\frac{\pi}{7}\right) + \tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)$, we need to take advantage of the identity involving the tangent of angles that sum up to $\pi$.
Given a triangle with angles $\frac{\pi}{7}$, $\frac{2\pi}{7}$, and $\frac{4\pi}{7}$, we know:

• $\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{4\pi}{7}\right)\tan\left(\frac{\pi}{7}\right)=\tan\left(\frac{\pi}{7}\right)$
• Sum of these angles is $\pi$, so the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ when $A + B + C = \pi$ can be used:

Therefore:

$(\tan\left(\frac{2\pi}{7}\right)+\tan\left(\frac{4\pi}{7}\right)+\tan\left(\frac{\pi}{7}\right))/\tan\left(\frac{\pi}{7}\right)=1$

So:

• $\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{4\pi}{7}\right) + \tan\left(\frac{4\pi}{7}\right)\tan\left(\frac{\pi}{7}\right) + \tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)=-(\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7})$
• Replacing the expression with $1$ derived from the product of the tangents:
• The expression simplifies to $\boxed{-7}$.

In essence, the property of tangent in triangles and the identity utilized allowed the expression to simplify elegantly.