If `cot^(-1)(n/(pi))>(pi)/6, n in N`, then the maximum value of n is :

Answer

To solve the inequality \( \cot^{-1}\left(\frac{n}{\pi}\right) > \frac{\pi}{6} \) where \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Understand the inequality
We start with the inequality:
\[ \cot^{-1}\left(\frac{n}{\pi}\right) > \frac{\pi}{6} \]

### Step 2: Apply the cotangent function
Since the cotangent function is decreasing, we can apply the cotangent to both sides of the inequality:
\[ \frac{n}{\pi} < \cot\left(\frac{\pi}{6}\right) \]

### Step 3: Calculate \( \cot\left(\frac{\pi}{6}\right) \)
We know that:
\[ \cot\left(\frac{\pi}{6}\right) = \frac{1}{\tan\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \]

### Step 4: Substitute back into the inequality
Now we substitute this value back into our inequality:
\[ \frac{n}{\pi} < \sqrt{3} \]

### Step 5: Solve for \( n \)
Multiplying both sides by \( \pi \) gives us:
\[ n < \pi \sqrt{3} \]

### Step 6: Approximate \( \pi \sqrt{3} \)
Using \( \pi \approx \frac{22}{7} \) and \( \sqrt{3} \approx 1.732 \):
\[ \pi \sqrt{3} \approx \frac{22}{7} \times 1.732 \approx \frac{22 \times 1.732}{7} \approx \frac{38.064}{7} \approx 5.43 \]

### Step 7: Determine the maximum value of \( n \)
Since \( n \) must be a natural number, the maximum value of \( n \) that satisfies \( n < 5.43 \) is:
\[ n = 5 \]

### Conclusion
Thus, the maximum value of \( n \) is:
\[ \boxed{5} \]