Question

Write the value of $\tan \left( 2{{\tan }^{-1}}\dfrac{1}{5} \right)$.

Answer 1

Hint: In the given expression first expand the $2{{\tan }^{-1}}\dfrac{1}{5}$using the $2{{\tan }^{-1}}x$formulae take the value of $x=\dfrac{1}{5}$and substitute it in the formulae and we know that $\tan \left( {{\tan }^{-1}}\theta \right)=\theta$. By using these two formulas we will get the value of $\tan \left( 2{{\tan }^{-1}}\dfrac{1}{5} \right)$.

Complete step-by-step answer:

Given that we have to find the value of $\tan \left( 2{{\tan }^{-1}}\dfrac{1}{5} \right)$
We know that the formula for $2{{\tan }^{-1}}x$is given by $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$
Here the value of $x$ is $x=\dfrac{1}{5}$
Now apply the above formula to the expression we will get
$=\tan \left[ {{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{5}}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right) \right]$. . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$=\tan \left[ {{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{1-\dfrac{1}{25}} \right) \right]$ . . . . . . . . . . . . . . . . . . . . . . . . . . .. .(2)
$=\tan \left[ {{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{\dfrac{24}{25}} \right) \right]$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .(3)
$=\tan \left[ {{\tan }^{-1}}\left( \dfrac{2}{5}\times \dfrac{25}{24} \right) \right]$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
$=\tan \left[ {{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right]$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(5)
$=\dfrac{5}{12}$
So we get the value of \tan \left( 2{{\tan }^{-1}}\dfrac{1}{5} \right)$$$$=\dfrac{5}{12}.

Note: The inverse function tan exists for every real number. For every $x\in \left( -\infty ,\infty \right)$it is called the inverse tangent function or arc tangent function. So if ${{\tan }^{-1}}x=\theta$ then $\tan \theta =x$. The domain of ${{\tan }^{-1}}x$is set of all real numbers or else denoted by R and range is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.