Question

${\tan }^{-1}\left(\frac{1}{4}\right)+2{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{6}\right)+{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{4}$.

Answer 1

$\tan \left(\frac{\frac{1}{x}+\frac{1}{4}}{1-\frac{1}{x}\times \frac{1}{4}}\right)+{\tan }^{-1}\left\{\frac{2\times \frac{1}{5}}{1-{\left(\frac{1}{5}\right)}^2}\right\}+{\tan }^{-1}\frac{1}{6}=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{4+x}{4x-1}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)+{\tan }^{-1}\frac{1}{6}=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{4+x}{4x-1}\right)+{\tan }^{-1}\left\{\frac{\frac{6}{12}+\frac{1}{6}}{1-\frac{5}{12}\times \frac{1}{6}}\right\}=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{4+x}{4x-1}\right)+{\tan }^{-1}\left(\frac{30+12}{72-5}\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{4+x}{4x-1}\right)+{\tan }^{-1}\left(\frac{42}{67}\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left\{\frac{\frac{4+x}{4x-1}+\frac{42}{67}}{1-\left(\frac{4+x}{4x-1}\right)\left(\frac{42}{67}\right)}\right\}=\frac{\pi }{4}$

$\Rightarrow \frac{67(4+x)+42(4x-1)}{(4x-1)67-42(4+x)}=\tan \frac{\pi }{4}$

$\Rightarrow \frac{268+67x+167x-42}{268x-67-168-42x}=1$

$\Rightarrow 267+67x+168x-42=268x-67-168-42x$

$\Rightarrow 9x=-461$

$\Rightarrow x=\frac{-461}{9}$