For n in N , if cot -1 3+cot -1 4+cot -1 5+cot 1 n =π/4 , then n is equal to

Mathematics · Inverse Trigonometric Functions

JEE Main 2024 — 6 April, Shift 1 — Question 28

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

Numerical answer — enter this value.

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

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

$\tan ^{-1}\left(\frac{46}{48}\right)+\tan ^{-1} \frac{1}{\mathrm{n}}=\frac{\pi}{4}$

$\tan ^{-1}\left(\frac{23}{24}\right)+\tan ^{-1} \frac{1}{\mathrm{n}}=\frac{\pi}{4}$

$\tan ^{-1} \frac{1}{\mathrm{n}}=\tan ^{-1} 1-\tan ^{-1} \frac{23}{24}$

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

$\tan ^{-1} \frac{1}{\mathrm{n}}=\tan ^{-1}\left(\frac{\frac{1}{24}}{\frac{47}{24}}\right)$

$\tan ^{-1} \frac{1}{\mathrm{n}}=\tan ^{-1} \frac{1}{47}$

$\mathrm{n}=47$

Answer key and solution verified before publishing.

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