We recognize that the given expression follows the tangent addition formula:

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

By comparing, we see that:

• $A = \frac{\pi}{18}$
• $B = \frac{\pi}{9}$

Since $\frac{\pi}{9} = 2 \times \frac{\pi}{18}$, we add them:

$A + B = \frac{\pi}{18} + \frac{\pi}{9} = \frac{\pi}{18} + \frac{2\pi}{18} = \frac{3\pi}{18} = \frac{\pi}{6}$

Thus, applying the formula:

$\tan \left(\frac{\pi}{18} + \frac{\pi}{9}\right) = \tan \frac{\pi}{6}$

Since $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, we conclude:

$\frac{\tan \frac{\pi}{18} + \tan \frac{\pi}{9}}{1 - \tan \frac{\pi}{18} \tan \frac{\pi}{9}} = \frac{1}{\sqrt{3}}$

Final Answer:

$\frac{1}{\sqrt{3}}$

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Would you like a breakdown of the derivation? Here are five related questions to explore further:

1. How do you derive the tangent addition formula?
2. What are the values of $\tan \frac{\pi}{18}$ and $\tan \frac{\pi}{9}$?
3. Can this problem be solved using sine and cosine identities?
4. What happens if the formula is used for subtraction instead of addition?
5. How can we use the tangent addition formula for angles not commonly found in trigonometric tables?

Tip: Memorizing fundamental trigonometric values for special angles (like $30^\circ, 45^\circ,$ and $60^\circ$) helps simplify expressions quickly!