Question

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

• A. 43
• B. 47
• C. 49
• D. 51

Answer 1

Answer: (B)

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