Question

Consider $x=4{\tan }^{-1}\left(\frac{1}{5}\right),y={\tan }^{-1}\left(\frac{1}{70}\right)$ and $z={\tan }^{-1}\left(\frac{1}{99}\right)$.

What is $x-y$ equal to?

• A. ${\tan }^{-1}\left(\frac{828}{845}\right)$
• B. ${\tan }^{-1}\left(\frac{8287}{8450}\right)$
• C. ${\tan }^{-1}\left(\frac{8281}{8450}\right)$
• D. ${\tan }^{-1}\left(\frac{8287}{8471}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{8281}{8450}\right)$

$x=4{\tan }^{-1}\left(\frac{1}{5}\right)=2{\tan }^{-1}\left(\frac{1}{5}\right)+2{\tan }^{-1}\left(\frac{1}{5}\right)=2{\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5}\times \frac{1}{5}}\right)=2{\tan }^{-1}\left(\frac{5}{12}\right)={\tan }^{-1}\left(\frac{120}{119}\right)$

$\therefore$ $x-y={\tan }^{-1}\left(\frac{120}{119}\right)-{\tan }^{-1}\left(\frac{1}{70}\right)$

$={\tan }^{-1}\left(\frac{\frac{120}{119}-\frac{1}{70}}{1+\frac{120}{119}\times \frac{1}{70}}\right)$

$={\tan }^{-1}\left(\frac{8281}{8450}\right)$

Hence, $x-y={\tan }^{-1}\left(\frac{8281}{8450}\right)$