Answer
$\tan x=x+\frac{2}{3!}{{x}^{3}}+\cdots $
Work Step by Step
\[\begin{align}
& f\left( x \right)=\tan x,\text{ }c=0\text{ } \\
& \text{Calculate the derivatives: }f'\left( x \right),\text{ }f''\left( x \right),f'''\left( x \right),{{f}^{\left( 4 \right)}}\left( x \right) \\
& *f'\left( x \right)={{\sec }^{2}}x \\
& \\
& *f''\left( x \right)=2\sec x\left( \sec x\tan x \right) \\
& f''\left( x \right)=2{{\sec }^{2}}x\tan x \\
& \\
& *f'''\left( x \right)=2{{\sec }^{2}}x\left( {{\sec }^{2}}x \right)+\tan x\left( 4\sec x \right)\left( \sec x\tan x \right) \\
& f'''\left( x \right)=2{{\sec }^{4}}x+4{{\sec }^{2}}x{{\tan }^{2}}x \\
& \\
& *{{f}^{4}}\left( x \right)=8{{\sec }^{3}}x\left( \sec x\tan x \right)+4{{\sec }^{2}}x\left( 2\tan x \right)\left( {{\sec }^{2}}x \right) \\
& +{{\tan }^{2}}x\left( 8\sec x \right)\left( \sec x\tan x \right) \\
& {{f}^{4}}\left( x \right)=8{{\sec }^{4}}x\tan x+8{{\sec }^{4}}x\tan x+8{{\sec }^{2}}x{{\tan }^{3}}x \\
& {{f}^{4}}\left( x \right)=16{{\sec }^{4}}x\tan x+8{{\sec }^{2}}x{{\tan }^{3}}x \\
& \\
& \text{Calculate }{{f}^{n}}\left( c \right)\text{ at }c=0 \\
& f\left( 0 \right)=\tan \left( 0 \right)=0 \\
& f'\left( 0 \right)={{\sec }^{2}}\left( 0 \right)=1 \\
& f''\left( 0 \right)=2{{\sec }^{2}}\left( 0 \right)\tan \left( 0 \right)=0 \\
& f'''\left( 0 \right)=2{{\sec }^{4}}\left( 0 \right)+4{{\sec }^{2}}\left( 0 \right){{\tan }^{2}}\left( 0 \right)=2 \\
& {{f}^{4}}\left( 0 \right)=16{{\sec }^{4}}\left( 0 \right)\tan \left( 0 \right)+8{{\sec }^{2}}\left( 0 \right){{\tan }^{3}}\left( 0 \right)=0 \\
& \\
& \text{Using the definition of the Taylor series} \\
& \sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( c \right)}{n!}}{{\left( x-c \right)}^{n}} \\
& \text{For }c=0 \\
& =\frac{{{f}^{\left( 0 \right)}}\left( 0 \right)}{0!}\left( x-0 \right)+\frac{f'\left( 0 \right)}{1!}\left( x-0 \right)+\frac{f''\left( 0 \right)}{2!}{{\left( x-0 \right)}^{2}} \\
& \frac{f'''\left( 0 \right)}{3!}{{\left( x-0 \right)}^{3}}+\frac{{{f}^{\left( 4 \right)}}\left( 0 \right)}{4!}{{\left( x-0 \right)}^{4}}+\cdots \\
& \text{Substituting the previous result and simplifying} \\
& \tan x=\frac{0}{0!}+\frac{1}{1!}x+\frac{0}{2!}{{x}^{2}}+\frac{2}{3!}{{x}^{3}}+\frac{0}{4!}{{x}^{4}}+\cdots \\
& \tan x=x+\frac{2}{3!}{{x}^{3}}+\cdots \\
\end{align}\]
