To solve the problem, we need to evaluate the expression:

\[\tan \frac{20\pi}{21} - \tan \frac{2\pi}{7} + \sqrt{3}\tan \frac{2\pi}{7} \tan \frac{20\pi}{21}\]

We can make progress by utilizing the following trigonometric identity:

\[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]

In this case, suppose:

• A = \frac{20\pi}{21}
• B = \frac{2\pi}{7}\ (or equivalently B = \frac{6\pi}{21}, since \frac{2\pi}{7} = \frac{6\pi}{21})

Then compute A + B:

A + B = \frac{20\pi}{21} + \frac{6\pi}{21} = \frac{26\pi}{21}\

We know that:

\[\tan \left(\frac{26\pi}{21}\right)\] can be simplified using the periodicity of tangent function:

\[\tan \left(\frac{26\pi}{21}\right) = \tan \left(\pi + \frac{5\pi}{21}\right) = \tan \left(\frac{5\pi}{21}\right)\]

Since \(\pi + x\) does not change the tangent function's value, we have:

\[\frac{\tan \frac{20\pi}{21} + \tan \frac{2\pi}{7}}{1 - \tan \frac{20\pi}{21} \tan \frac{2\pi}{7}} = \tan \frac{5\pi}{21}\]

Let's equate it to given expression, factoring:

\[\tan \frac{20\pi}{21} - \tan \frac{2\pi}{7} + \sqrt{3}\tan \frac{2\pi}{7} \tan \frac{20\pi}{21} = \frac{\tan \frac{20\pi}{21} + \tan \frac{2\pi}{7}}{1 - \tan \frac{20\pi}{21} \tan \frac{2\pi}{7}}\]

Thus:

The expression simplifies to:

\[\sqrt{3} = \tan \frac{5\pi}{21}.\]

Therefore, the given equation holds true if

The correct option is:

-\sqrt{3}, which simplifies the equation correctly.