For n in N,if cot ^{-1} 3+cot ^{-1} 4+cot ^{- | vedclass

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(D) Given the equation: $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$.Converting to $	an ^{-1}$ using $\cot ^{-1} x = 	an ^{-1

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$47$

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(D) Given the equation: $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$.Converting to $	an ^{-1}$ using $\cot ^{-1} x = 	an ^{-1} (1/x)$ for $x > 0$:$	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{4}+	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{n}=\frac{\pi}{4}$.First,combine $	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{4} = 	an ^{-1} \left(\frac{1/3+1/4}{1-(1/3)(1/4)}ight) = 	an ^{-1} \left(\frac{7/12}{11/12}ight) = 	an ^{-1} \frac{7}{11}$.Now,add $	an ^{-1} \frac{1}{5}$: $	an ^{-1} \frac{7}{11}+	an ^{-1} \frac{1}{5} = 	an ^{-1} \left(\frac{7/11+1/5}{1-(7/11)(1/5)}ight) = 	an ^{-1} \left(\frac{46/55}{48/55}ight) = 	an ^{-1} \frac{46}{48} = 	an ^{-1} \frac{23}{24}$.So,$	an ^{-1} \frac{23}{24} + 	an ^{-1} \frac{1}{n} = \frac{\pi}{4}$.$	an ^{-1} \frac{1}{n} = 	an ^{-1} 1 - 	an ^{-1} \frac{23}{24} = 	an ^{-1} \left(\frac{1-23/24}{1+(1)(23/24)}ight) = 	an ^{-1} \left(\frac{1/24}{47/24}ight) = 	an ^{-1} \frac{1}{47}$.Therefore,$n = 47$.

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

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