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 $x-y+z$ equal to?

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

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

$x=2[{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)]=2{\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5}\times \frac{1}{5}}\right)={\tan }^{-1}\left(\frac{120}{119}\right)\Rightarrow x-y={\tan }^{-1}\left(\frac{\frac{120}{119}-\frac{1}{70}}{1+\frac{120}{119}\times \frac{1}{70}}\right)={\tan }^{-1}\left(\frac{8281}{8450}\right)\Rightarrow x-y+z={\tan }^{-1}\left(\frac{8281}{8450}\right)+{\tan }^{-1}\left(\frac{1}{99}\right)={\tan }^{-1}\left(\frac{828269}{828269}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$