Question

How do you find the exact value of $\arctan (\tan x)$ ?

Answer 1

$\arctan (\tan x)$ is nothing but ${{\tan }^{-1}}\left( \tan x \right)$ . The values of ${{\tan }^{-1}}\left( \tan x \right)$ changes as the quadrant changes. To graph it, we need to know it’s range and it’s domain as well. The domain of ${{\tan }^{-1}}\left( x \right)$ is R which means any real value of $x$ can be substituted. It’s range is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ . This is called the principal range of ${{\tan }^{-1}}\left( x \right)$. Let us use this information to see how the graph of ${{\tan }^{-1}}\left( \tan x \right)$ at different places and graph it.

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Complete step by step solution:

Let us assume that ${{\tan }^{-1}}\left( \tan x \right)=y$. So it would become the following:

$\tan^{-1}(x)=y$

$\Rightarrow \tan x=\tan y$

$\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ is called the principal range of ${{\tan }^{-1}}\left( x \right)$. Whatever value we get upon substituting any $x$ , that value must lie between $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.

First case : Let us randomly substitute $x=0$ and see what value of $y$ we obtain.

$\tan (x) = \tan (y)$

$\Rightarrow \tan \left( 0 \right)=\tan y$

$\Rightarrow \tan y=0$

$\Rightarrow y=0$

$0$ lies in between $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.$\left( 0,0 \right)$ is a point in the graph.

Second case : Let us take a random value of $x$ from it’s domain. $x=\dfrac{\pi }{4}$ and see what value of $y$ we obtain.

$\tan (\dfrac{\pi }{4}) = \tan (y)$

$\Rightarrow 1=\tan y$

$\Rightarrow \tan y=1$

$\Rightarrow y=\dfrac{\pi }{4}$

$\dfrac{\pi }{4}$ lies in between $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.$\left( \dfrac{\pi }{4},\dfrac{\pi }{4} \right)$ is a point in the graph.

Third case : Let us take another random value of $x$ from it’s domain. $x=\dfrac{-\pi }{4}$ and see what value of $y$ we obtain.

$\tan (\dfrac{\pi }{4}) = \tan (y)$

$\Rightarrow -1=\tan y$

$\Rightarrow \tan y=-1$

$y=-\dfrac{\pi }{4}$

Value of $y$ would be $\dfrac{-\pi }{4}$not $\dfrac{3\pi }{4}$ since $\dfrac{3\pi }{4}$ is not in the range of ${{\tan }^{-1}}\left( x \right)$.

So when $x$ is in it’s domain i.e in $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$, we are getting $x=y$. So this would be the graph of our function ${{\tan }^{-1}}\left( \tan x \right)$from $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. At $x=\pm \dfrac{\pi }{2}$ , ${{\tan }^{-1}}x$ would be a discontinuous function.

We can generalize this over the set of different domains.

$\Rightarrow {{\tan }^{-1}}\left( \tan y \right)=x-n\pi$ for $\left( 2n-1 \right)\dfrac{\pi }{2}$ < $x$ < $\left( 2n+1 \right)\dfrac{\pi }{2},n\in Z$

Graph:

Note: Please do not get confused as there are numbers on the graph. The angles are in radians here. $\pi$ is nothing but $3.14$ radians. Also, we should remember the domain and range of all the inverse trigonometric functions so as to be able to graph them. This is a very important chapter. There are a lot of formulae to remember along with their specifications. There is a huge scope for calculation errors. A lot of practice is required for this chapter.