Question

How do you evaluate $\tan \left( {4{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right)} \right)$?

Answer 1

In the given question we have to find the value of the given expression which is a trigonometric function, we will evaluate by first considering the inverse function inside as a variable, then by using the identity $\tan \left( {{{\tan }^{ - 1}}x} \right) = x$, we will get the tan value of the variable, then we will use the double angle identity i.e., $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$, and by substituting the values in the identity we will get the required value.

Complete step by step solution:
Given function is $\tan \left( {4{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right)} \right)$,
Let us consider $a = {\tan ^{ - 1}}\dfrac{1}{5}$,
Apply tan on both sides we get,
$\Rightarrow \tan a = \tan \left( {{{\tan }^{ - 1}}\dfrac{1}{5}} \right)$,
Now using the identity $\tan \left( {{{\tan }^{ - 1}}x} \right) = x$, we get,
$\Rightarrow \tan a = \dfrac{1}{5}$,
Now using double angle identity i.e., $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$, we get,
$\Rightarrow \tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$,
We know that $\tan a = \dfrac{1}{5}$, by substituting the value in the identity, we get,
$\Rightarrow \tan 2a = \dfrac{{2\left( {\dfrac{1}{5}} \right)}}{{1 - {{\left( {\dfrac{1}{5}} \right)}^2}}}$,
Now simplifying we get,
$\Rightarrow \tan 2a = \dfrac{{\dfrac{2}{5}}}{{1 - \dfrac{1}{{25}}}}$,
Now taking L.C.M in the denominator we get,
$\Rightarrow \tan 2a = \dfrac{{\dfrac{2}{5}}}{{\dfrac{{25 - 1}}{{25}}}}$,
Again simplifying we get,
$\Rightarrow \tan 2a = \dfrac{{\dfrac{2}{5}}}{{\dfrac{{24}}{{25}}}}$,
Now further simplifying, we get,
$\Rightarrow \tan 2a = \dfrac{1}{{\dfrac{{12}}{5}}}$,
Now taking denominator of the denominator to the numerator we get,
$\Rightarrow \tan 2a = \dfrac{5}{{12}}$,
Again the given expression will be, $\tan 4a = \dfrac{{2\tan 2a}}{{1 - {{\tan }^2}2a}}$, we know that the value of $\tan 2a = \dfrac{5}{{12}}$, by substituting the value in the identity we get,
$\Rightarrow \tan 4a = \dfrac{{2\left( {\dfrac{5}{{12}}} \right)}}{{1 - {{\left( {\dfrac{5}{{12}}} \right)}^2}}}$,
Now simplifying we get,
$\Rightarrow \tan 4a = \dfrac{{\dfrac{5}{6}}}{{1 - \dfrac{{25}}{{144}}}}$,
Now taking L.C.M in the denominator we get,
$\Rightarrow \tan 4a = \dfrac{{\dfrac{5}{6}}}{{\dfrac{{144 - 25}}{{144}}}}$,
Now simplifying we get,
$\Rightarrow \tan 4a = \dfrac{{\dfrac{5}{6}}}{{\dfrac{{119}}{{144}}}}$,
Now simplifying we get,
$\Rightarrow \tan 4a = \dfrac{5}{{\dfrac{{119}}{{24}}}}$,
Again simplifying we get,
$\Rightarrow \tan 4a = \dfrac{{5 \times 24}}{{119}}$,
Now multiplying we get,
$\Rightarrow \tan 4a = \dfrac{{120}}{{119}}$ ,
Now we know that $a = {\tan ^{ - 1}}\dfrac{1}{5}$, by substituting the value in the above result we get,
$\Rightarrow \tan 4\left( {{{\tan }^{ - 1}}\dfrac{1}{5}} \right) = \dfrac{{120}}{{119}}$,
So, the value of the expression is $\dfrac{{120}}{{119}}$,

$\therefore$The value of the given expression which is $\tan \left( {4{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right)} \right)$ will be equal to $\dfrac{{120}}{{119}}$.

Note: The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself. Some of the important formulas or identities are:
$\sin 2x = 2\sin x\cos x$,
$\cos 2x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1 = {\cos ^2}x - {\sin ^2}x$,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$.