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: $n = 47$