1. Tardigrade
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  4. For n ∈ N, if tan -1 (1/3)+ tan -1 (1/4)+ tan -1 (1/5)+ tan -1 (1/ n )=(π/4) then n is equal to

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

Inverse Trigonometric Functions

Solution:

We have, $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{4}=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{12}}\right)=\tan ^{-1}\left(\frac{7}{11}\right) $
Again, $\tan ^{-1} \frac{7}{11}+\tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{\frac{7}{11}+\frac{1}{5}}{1-\frac{7}{55}}\right)=\tan ^{-1}\left(\frac{46}{48}\right)=\tan ^{-1} \frac{23}{24}$
$\therefore \tan ^{-1} \frac{1}{ n }=\tan ^{-1} 1-\tan ^{-1} \frac{23}{24}=\tan ^{-1}\left(\frac{1-\frac{23}{24}}{1+\frac{23}{24}}\right)=\tan ^{-1}\left(\frac{1}{47}\right) \Rightarrow n =47 $