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

<p>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 <strong></strong><strong></strong>.</p> <p>Given the equation:</p> <p>$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $</p> <p>we can use the identity for the sum of inverse cotangents. Starting with the first two terms:</p> <p>$ \cot^{-1} 3 + \cot^{-1} 4 = \cot^{-1}\left(\frac{3 \times 4 - 1}{3 + 4}\right) = \cot^{-1}\left(\frac{11}{7}\right) $</p> <p>Now, adding the third term:</p> <p>$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}(5) $</p> <p>we apply the identity again:</p> <p>$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}\left(\frac{n \times 5 - 1}{5 + n}\right) = \frac{\pi}{4} $</p> <p>Rewriting this to isolate the sum of the terms, we proceed as follows:</p> <p>$ \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} $</p> <p>This simplifies to:</p> <p>$ \frac{11}{7} \left(\frac{5n-1}{5+n}\right) - 1 = \frac{11}{7} + \frac{5n-1}{5+n} $</p> <p>Solving the equation:</p> <p>$ \frac{55n - 11}{5 + n} - 1 = \frac{11}{7} + \frac{5n - 1}{5 + n} $</p> <p>Further simplification yields:</p> <p>$ 55n - 11 - 35 - 7n = 55 + 11n + 35n - 7 $</p> <p>Bringing the terms together, we get:</p> <p>$ 48n - 46 = 48 $</p> <p>Therefore:</p> <p>$ 2n = 94 $</p> <p>So finally:</p> <p>$ n = 47 $</p>