Question

Prove
$4{\tan }^{-1}\left(\frac{1}{5}\right)-{\tan }^{-1}\left(\frac{1}{70}\right)+{\tan }^{-1}\left(\frac{1}{99}\right)=\frac{\pi }{4}$.

Answer 1

To be proved : $4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\left(\frac{1}{70}\right)+{\tan }^{-1}\left(\frac{1}{99}\right)=\frac{\pi }{4}$

We know that $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

So, $2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{2\times \frac{1}{5}}{1-{\left(\frac{1}{5}\right)}^2}\right)={\tan }^{-1}\left(\frac{5}{12}\right)$

$LHS:4{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{99}-{\tan }^{-1}\frac{1}{70}$

$=(2{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{70})+(2{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{70})+(2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{99})$

$=({\tan }^{-1}\frac{5}{12}-{\tan }^{-1}\frac{1}{70})+({\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{1}{99})$

$={\tan }^{-1}\left(\frac{\frac{5}{12}-\frac{1}{70}}{1+\left(\frac{5}{12}\right)\left(\frac{1}{70}\right)}\right)+{\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{1}{99}}{1-\left(\frac{5}{12}\right)\left(\frac{1}{99}\right)}\right)$

$(\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right))$

$={\tan }^{-1}\left(\frac{2}{5}\right)+{\tan }^{-1}\left(\frac{3}{7}\right)$

$={\tan }^{-1}\left(\frac{\frac{2}{5}+\frac{3}{7}}{1-\frac{6}{35}}\right)={\tan }^{-1}\left(1\right)$

$=\frac{\pi }{4}$