How do you calculate $\left( {{\tan }^{-1}}\left( 2 \right) \right)$ ?

Hint: In this problem we have to calculate the $\left( {{\tan }^{-1}}\left( 2 \right) \right)$. We will use $\dfrac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+.....$ formula for this problem. We will integrate the above equation and simplify the obtained equation. We know that $\int{\dfrac{1}{1+{{x}^{2}}}}dx={{\tan }^{-1}}x+C$ . Here we need to calculate the value of ${{\tan }^{-1}}2$, which does not satisfy the condition $-1\le x\le 1$. So, we will use the trigonometric formula ${{\tan }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{1}{x} \right)$. So here we will get $x=\dfrac{1}{2}$ which satisfies $-1\le x\le 1$. Now we can use the integration value and use the Tylor series to get the result.

Formula used:
1. $\dfrac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+.....$
2. $\int{\dfrac{1}{1+{{x}^{2}}}}dx={{\tan }^{-1}}x+C$
3. ${{\tan }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{1}{x} \right)$
4. $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$

Complete Step by Step Procedure:
Given that, $\left( {{\tan }^{-1}}\left( 2 \right) \right) $
Consider the equation
$\dfrac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+.....$
Integrating the above equation with respect to $x$, then we will get
 $\int{\dfrac{1}{1+{{x}^{2}}}dx=\int{\left( 1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+..... \right)dx}}$
We know that $\int{\dfrac{1}{1+{{x}^{2}}}}dx={{\tan }^{-1}}x+C$ and using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ in the above equation, then we will get
${{\tan }^{-1}}\left( x \right)=x-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{7}}}{7}+.....$
The above formula can be used when $-1\le x\le 1$. But in this problem, we have to calculate the value of ${{\tan }^{-1}}2$, where $x=2$ and $x>2$. So here we can’t use the above formula directly.
We have trigonometric equation i.e.,
${{\tan }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{1}{x} \right)$
Substituting $x=2$ in the above equation, then we will get
$\Rightarrow {{\tan }^{-1}}(2)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{1}{2} \right)....\left( \text{i} \right)$
Now we will calculate the value of ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)$ by using the above integration value.
Substituting the $x=\dfrac{1}{2}$in the integration value, then we will get
${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{1}{2}-\dfrac{1}{8\times 4}+\dfrac{1}{32\times 5}-\dfrac{1}{128\times 7}+....$
Simplifying the above equation, then we will get
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2} \right)=0.463467$
Now substitute ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=0.463467$in the equation $\left( \text{i} \right)$, then we will get
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( 2 \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( 2 \right)=\dfrac{\pi }{2}-0.463467 \\
\end{align}$
Substituting the value of $\pi =3.14$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( 2 \right)=1.52-0.463467 \\
 & \Rightarrow {{\tan }^{-1}}\left( 2 \right)=1.107329 \\
\end{align}$
In the above equation we have the result in the radians. To convert the radians into degrees we will multiply above value with $\dfrac{180}{\pi }$,then

$\begin{align}
  & {{\tan }^{-1}}\left( 2 \right)=\dfrac{180}{\pi }\times 1.107329 \\

& {{\tan }^{-1}}\left( 2 \right)=63.4452{}^\circ \\
\end{align}$

Note:
In the above problem, we have calculated the value of ${{\tan }^{-1}}x$ from the Tylor series. But it can be easier to use a scientific calculator to find the value of ${{\tan }^{-1}}x$. Because there is a drawback with this series. The series can only give the value of ${{\tan }^{-1}}x$ where $-1\le x\le 1$. For other values, we need to use some trigonometric identities and solve the problem according to the previous step.