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 \[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\].