For $ n \in \mathbb{N} $, if $ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} $, then $ n $ is equal to .

Given the equation:

$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $

we can use the identity for the sum of inverse cotangents. Starting with the first two terms:

$ \cot^{-1} 3 + \cot^{-1} 4 = \cot^{-1}\left(\frac{3 \times 4 - 1}{3 + 4}\right) = \cot^{-1}\left(\frac{11}{7}\right) $

Now, adding the third term:

$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}(5) $

we apply the identity again:

$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}\left(\frac{n \times 5 - 1}{5 + n}\right) = \frac{\pi}{4} $

Rewriting this to isolate the sum of the terms, we proceed as follows:

$ \cot^{-1}\left( \frac{\left(\frac{11}{7} \times \frac{5n-1}{5+n} - 1\right)}{\left(\frac{11}{7} + \frac{5n-1}{5+n} \right)} \right) = \frac{\pi}{4} $

This simplifies to:

$ \frac{11}{7} \left(\frac{5n-1}{5+n}\right) - 1 = \frac{11}{7} + \frac{5n-1}{5+n} $

Solving the equation:

$ \frac{55n - 11}{5 + n} - 1 = \frac{11}{7} + \frac{5n - 1}{5 + n} $

Further simplification yields:

$ 55n - 11 - 35 - 7n = 55 + 11n + 35n - 7 $

Bringing the terms together, we get:

$ 48n - 46 = 48 $

Therefore:

$ 2n = 94 $

So finally:

$ n = 47 $