Find the value of
`tan^(-1)("tan"(7pi)/(6))+ cos^(-1)("cos"(7pi)/(6))`

Answer

To solve the expression \( \tan^{-1}(\tan(7\pi/6)) + \cos^{-1}(\cos(7\pi/6)) \), we can follow these steps:

### Step 1: Analyze the angles
The angle \( 7\pi/6 \) is in the third quadrant. We need to find the equivalent angles that lie within the principal ranges of the inverse functions.

### Step 2: Find the equivalent angle for \( \tan^{-1}(\tan(7\pi/6)) \)
The function \( \tan^{-1}(x) \) has a range of \( (-\pi/2, \pi/2) \). Since \( 7\pi/6 \) is greater than \( \pi/2 \), we can express it as:
\[ 7\pi/6 = \pi + \pi/6 \]
Thus, we can write:
\[ \tan(7\pi/6) = \tan(\pi + \pi/6) = \tan(\pi/6) \]
So, we have:
\[ \tan^{-1}(\tan(7\pi/6)) = \tan^{-1}(\tan(\pi/6)) = \pi/6 \]

### Step 3: Find the equivalent angle for \( \cos^{-1}(\cos(7\pi/6)) \)
The function \( \cos^{-1}(x) \) has a range of \( [0, \pi] \). Since \( 7\pi/6 \) is greater than \( \pi \), we can express it as:
\[ 7\pi/6 = \pi + \pi/6 \]
Thus, we can write:
\[ \cos(7\pi/6) = \cos(\pi + \pi/6) = -\cos(\pi/6) \]
So, we have:
\[ \cos^{-1}(\cos(7\pi/6)) = \cos^{-1}(-\cos(\pi/6)) \]
Using the property of the cosine function, we know:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
Thus:
\[ \cos^{-1}(-\cos(\pi/6)) = \pi - \cos^{-1}(\cos(\pi/6)) = \pi - \pi/6 = 5\pi/6 \]

### Step 4: Combine the results
Now we can combine the results from Step 2 and Step 3:
\[ \tan^{-1}(\tan(7\pi/6)) + \cos^{-1}(\cos(7\pi/6)) = \frac{\pi}{6} + \frac{5\pi}{6} \]
Calculating this gives:
\[ \frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi \]

### Final Answer
Thus, the value of \( \tan^{-1}(\tan(7\pi/6)) + \cos^{-1}(\cos(7\pi/6)) \) is:
\[ \boxed{\pi} \]