Question

Prove that ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{4}$

Answer 1

We know that, ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Replace $xlog1/5$ and $ylog1/7$

${\tan }^{-1}1/5+{\tan }^{-1}1/7={\tan }^{-1}\frac{1/5+1/7}{1-1/5\times 1/7}={\tan }^{-1}\left(6/17\right)$

Similarly replace $2log1/3$ and $y$ $log1/8$

${\tan }^{-1}1/3+{\tan }^{-1}1/8-{\tan }^{-1}\frac{1/3+1/2}{1-1/3\times 1/8}={\tan }^{-1}\left(11/23\right)$

LHS$\Rightarrow {\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{8}$

$=({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7})+{\tan }^{-1}(\frac{1}{3}+{\tan }^{-}\frac{1}{8})$

$={\tan }^{-1}\left(6/17\right)+{\tan }^{-1}\left(11/23\right)$

$={\tan }^{-1}\left[\frac{6/17+11/23}{1-6/17\times 11/13}\right]$

$={\tan }^{-1}\frac{138+187/391}{391-66/391}$

$={\tan }^{-1}\frac{325}{391}\times \frac{391}{325}={\tan }^{-1}1$

$=ta{n}^{-1}$

$={\tan }^{-1}\left(\tan \frac{\pi }{4}\right)\left[\begin{array}{c}\tan 45=7\\ \tan \pi /4=7\end{array}\right]$

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

$=$RHS.