Evaluate: `tan^(-1)(tan(5pi)/6)+cos^(-1){cos((13pi)/6)}`

Explanation

To evaluate the expression \( \tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6)) \), we will follow these steps:

### Step 1: Evaluate \( \tan^{-1}(\tan(5\pi/6)) \)
The function \( \tan^{-1}(x) \) is defined for \( x \) in the range \(-\frac{\pi}{2} < x < \frac{\pi}{2}\).

1. First, we need to find \( \tan(5\pi/6) \).
\[ \tan(5\pi/6) = \tan(\pi - \frac{\pi}{6}) = -\tan(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \]

2. Now, we find \( \tan^{-1}(-\frac{1}{\sqrt{3}}) \). Since \(-\frac{1}{\sqrt{3}}\) is within the range of \( \tan^{-1} \), we can directly find:
\[ \tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6} \]

### Step 2: Evaluate \( \cos^{-1}(\cos(13\pi/6)) \)
The function \( \cos^{-1}(x) \) is defined for \( x \) in the range \( 0 \leq x \leq \pi \).

1. First, we need to find \( \cos(13\pi/6) \).
\[ 13\pi/6 = 2\pi + \frac{\pi}{6} \quad (\text{since } 2\pi = 12\pi/6) \]
Thus,
\[ \cos(13\pi/6) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \]

2. Now, we find \( \cos^{-1}(\frac{\sqrt{3}}{2}) \). Since \( \frac{\sqrt{3}}{2} \) is within the range of \( \cos^{-1} \), we can directly find:
\[ \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} \]

### Step 3: Combine the results
Now we combine the results from Step 1 and Step 2:
\[ \tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6)) = -\frac{\pi}{6} + \frac{\pi}{6} = 0 \]

### Final Answer
Thus, the final result is:
\[ \boxed{0} \]