The value of `tan^(-1)("tan"(5pi)/6)+cos^(-1)("cos"(13pi)/6)` is:

Options

A. 0
B. `-1`
C. 2
D. 1

Answer

To solve the expression \( \tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6)) \), we will break it down step by step.

### Step 1: Simplify \( \tan(5\pi/6) \)
The angle \( 5\pi/6 \) can be expressed as:
\[ 5\pi/6 = \pi - \pi/6 \]
Using the tangent subtraction identity:
\[ \tan(\pi - \theta) = -\tan(\theta) \]
Thus:
\[ \tan(5\pi/6) = -\tan(\pi/6) \]
We know that:
\[ \tan(\pi/6) = \frac{1}{\sqrt{3}} \]
So:
\[ \tan(5\pi/6) = -\frac{1}{\sqrt{3}} \]

### Step 2: Find \( \tan^{-1}(\tan(5\pi/6)) \)
Since \( 5\pi/6 \) is in the second quadrant, where tangent is negative, we have:
\[ \tan^{-1}(\tan(5\pi/6)) = 5\pi/6 \]

### Step 3: Simplify \( \cos(13\pi/6) \)
The angle \( 13\pi/6 \) can be expressed as:
\[ 13\pi/6 = 2\pi + \pi/6 \]
Using the cosine periodicity:
\[ \cos(2\pi + \theta) = \cos(\theta) \]
Thus:
\[ \cos(13\pi/6) = \cos(\pi/6) \]
We know that:
\[ \cos(\pi/6) = \frac{\sqrt{3}}{2} \]

### Step 4: Find \( \cos^{-1}(\cos(13\pi/6)) \)
Since \( \pi/6 \) is in the first quadrant, we have:
\[ \cos^{-1}(\cos(13\pi/6)) = \pi/6 \]

### Step 5: Combine the results
Now we can combine the results from Steps 2 and 4:
\[ \tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6)) = \frac{5\pi}{6} + \frac{\pi}{6} \]
Adding these two fractions gives:
\[ \frac{5\pi}{6} + \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]

### Final Answer
Thus, the value of \( \tan^{-1}(\tan(5\pi/6)) + \cos^{-1}(\cos(13\pi/6)) \) is:
\[ \pi \]