Use a sum or difference formula to find the exact value of the following. \[ \frac{\tan \frac{19 \pi}{15}-\tan \frac{14 \pi}{15}}{1+\tan \frac{19 \pi}{15} \tan \frac{14 \pi}{15}} \]

Step 1 ：The given expression is in the form of the tangent difference formula, which is given by: \[\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\]

Step 2 ：We can rewrite the given expression as \(\tan(\frac{19\pi}{15} - \frac{14\pi}{15})\).

Step 3 ：Let's calculate the difference: \(\alpha = \frac{19\pi}{15}\), \(\beta = \frac{14\pi}{15}\), so the difference is \(\alpha - \beta = \frac{\pi}{3}\).

Step 4 ：The tangent of \(\frac{\pi}{3}\) is \(\sqrt{3}\).

Step 5 ：Final Answer: The exact value of the given expression is \(\boxed{\sqrt{3}}\).