Question

Evaluate :
$tan[2{tan}^{-1}\left(\frac{1}{5}\right)-\frac{\pi }{4}]$

Answer 1

$\tan [2{\tan }^{-1}\left(\frac{1}{5}\right)-\frac{\pi }{4}]$

We know that $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

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

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

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

Replace $x$ by $\frac{1}{5}$

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

$\Rightarrow 2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right)$

$\Rightarrow 2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{\frac{2}{5}}{\frac{25-1}{25}}\right)$

$\Rightarrow 2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)$

$\Rightarrow 2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{2}{24}\right)$

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

Substituting $2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\left(\frac{1}{12}\right)$ in

$\frac{\tan 2{\tan }^{-1}\left(\frac{1}{5}\right)-1}{1+\tan 2{\tan }^{-1}\left(\frac{1}{5}\right)}$ we get

$\Rightarrow \frac{\tan {\tan }^{-1}\left(\frac{1}{12}\right)-1}{1+\tan {\tan }^{-1}\left(\frac{1}{12}\right)}$

$=\frac{\frac{1}{12}-1}{1+\frac{1}{12}}$

$=\frac{\frac{1-12}{12}}{\frac{12+1}{12}}$

$=\frac{-11}{13}$