Find the expression that is equivalent to $\frac{\tan \frac{3 \pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{3 \pi}{4} \tan \frac{\pi}{6}}$. Choose the correct answer below. A. $\tan \frac{7 \pi}{12}$ B. $\tan \frac{\pi}{8}$ c. $\tan \frac{11 \pi}{12}$ D. $\tan \frac{3 \pi}{4}+\tan \frac{\pi}{6}$

Step 1 ：The given expression is \(\frac{\tan \frac{3 \pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{3 \pi}{4} \tan \frac{\pi}{6}}\)

Step 2 ：This expression is in the form of the tangent addition formula, which is \(\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\)

Step 3 ：Here, \(a = \frac{3\pi}{4}\) and \(b = \frac{\pi}{6}\)

Step 4 ：So, the expression is equivalent to \(\tan(a+b)\), which is \(\tan(\frac{3\pi}{4} + \frac{\pi}{6})\)

Step 5 ：Adding the values of a and b, we get \(\frac{3\pi}{4} + \frac{\pi}{6} = \frac{11\pi}{12}\)

Step 6 ：Final Answer: The expression is equivalent to \(\tan \frac{11 \pi}{12}\)

Step 7 ：So, the correct answer is \(\boxed{C. \tan \frac{11 \pi}{12}}\)