Question

Evaluate $\tan \left(2{\tan }^{-1}\frac{1}{5}\right)$.

Answer 1

given, $tan\left(2ta{n}^{-1}\frac{1}{5}\right)$ $[\because 2ta{n}^{-1}x=ta{n}^{-1}(\frac{2x}{1-x^2}\left)\right]$

$=tan\left[ta{n}^{-1}\frac{2\times \frac{1}{5}}{1-(\frac{1}{5}{)}^2}\right]$

$=tan\left[ta{n}^{-1}\frac{\frac{2}{5}}{\frac{24}{25}}\right]$

$=tan\left[ta{n}^{-1}\frac{5}{12}\right]$

$=5/12$