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 _____

Convert to $$\tan^{-1}$$:

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

Combine first two terms:

$$\tan^{-1}(\frac{1/3 + 1/4}{1 - 1/12}) = \tan^{-1}(\frac{7/12}{11/12}) = \tan^{-1}(\frac{7}{11})$$

Combine with third term:

$$\tan^{-1}(\frac{7/11 + 1/5}{1 - 7/55}) = \tan^{-1}(\frac{46/55}{48/55}) = \tan^{-1}(\frac{23}{24})$$

Solve for $$n$$:

$$\tan^{-1}(\frac{23}{24}) + \tan^{-1}(\frac{1}{n}) = \tan^{-1}(1)$$

$$\tan^{-1}(\frac{1}{n}) = \tan^{-1}(1) - \tan^{-1}(\frac{23}{24}) = \tan^{-1}(\frac{1 - 23/24}{1 + 23/24}) = \tan^{-1}(\frac{1/24}{47/24}) = \tan^{-1}(\frac{1}{47})$$

Correct Answer: 47