Evaluate : `cos^(-1)(cos""(13pi)/6)+tan^(-1)(tan""(7pi)/6)`.

Options

A. `pi/2`
B. `pi`
C. `(3pi)/9`
D. None of these

Answer

To evaluate the expression \( \cos^{-1}(\cos(13\pi/6)) + \tan^{-1}(\tan(7\pi/6)) \), we will proceed step by step.

### Step 1: Simplify \( \cos(13\pi/6) \)

First, we can simplify \( 13\pi/6 \):
\[ 13\pi/6 = 12\pi/6 + \pi/6 = 2\pi + \pi/6 \]
Since cosine has a periodicity of \( 2\pi \):
\[ \cos(13\pi/6) = \cos(\pi/6) \]

### Step 2: Evaluate \( \cos^{-1}(\cos(13\pi/6)) \)

Now we can evaluate \( \cos^{-1}(\cos(13\pi/6)) \):
\[ \cos^{-1}(\cos(13\pi/6)) = \cos^{-1}(\cos(\pi/6)) \]
Since \( \pi/6 \) is in the range of \( [0, \pi] \), we have:
\[ \cos^{-1}(\cos(\pi/6)) = \pi/6 \]

### Step 3: Simplify \( \tan(7\pi/6) \)

Next, we simplify \( 7\pi/6 \):
\[ 7\pi/6 = \pi + \pi/6 \]
Since tangent has a periodicity of \( \pi \):
\[ \tan(7\pi/6) = \tan(\pi/6) \]

### Step 4: Evaluate \( \tan^{-1}(\tan(7\pi/6)) \)

Now we evaluate \( \tan^{-1}(\tan(7\pi/6)) \):
\[ \tan^{-1}(\tan(7\pi/6)) = \tan^{-1}(\tan(\pi/6)) \]
Since \( \pi/6 \) is in the range of \( (-\pi/2, \pi/2) \), we have:
\[ \tan^{-1}(\tan(\pi/6)) = \pi/6 \]

### Step 5: Combine the results

Now we can combine the results:
\[ \cos^{-1}(\cos(13\pi/6)) + \tan^{-1}(\tan(7\pi/6) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Final Answer

Thus, the final answer is:
\[ \frac{\pi}{3} \]

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