Answer
$\sec \left( x \right)=1+\frac{1}{2!}{{x}^{2}}+\frac{5}{4!}{{x}^{4}}+\cdots $
Work Step by Step
\[\begin{align}
& f\left( x \right)=\sec 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 x\tan x \\
& \\
& *f''\left( x \right)=\sec x\left( {{\sec }^{2}}x \right)+\tan x\left( \sec x\tan x \right) \\
& f''\left( x \right)={{\sec }^{3}}x+\sec x{{\tan }^{2}}x \\
& \\
& *f'''\left( x \right)=\frac{d}{dx}\left[ {{\sec }^{3}}x \right]+\frac{d}{dx}\left[ \sec x{{\tan }^{2}}x \right] \\
& f'''\left( x \right)=3{{\sec }^{2}}x\left( \sec x\tan x \right)+\sec x\left( 2\tan x \right)\left( {{\sec }^{2}}x \right) \\
& +{{\tan }^{2}}x\left( \sec x\tan x \right) \\
& f'''\left( x \right)=3{{\sec }^{3}}x\tan x+2{{\sec }^{3}}x\tan x+\sec x{{\tan }^{3}}x \\
& f'''\left( x \right)=5{{\sec }^{3}}x\tan x+\sec x{{\tan }^{3}}x \\
& \\
& *{{f}^{4}}\left( x \right)=\frac{d}{dx}\left[ 5{{\sec }^{3}}x\tan x \right]+\frac{d}{dx}\left[ \sec x{{\tan }^{3}}x \right] \\
& {{f}^{4}}\left( x \right)=5{{\sec }^{3}}x{{\sec }^{2}}x+15{{\sec }^{2}}x\tan x\left( \sec x\tan x \right) \\
& +\sec x\left( 3{{\tan }^{2}}x \right)\left( {{\sec }^{2}}x \right)+{{\tan }^{3}}x\left( \sec x\tan x \right) \\
& {{f}^{4}}\left( x \right)=5{{\sec }^{5}}x+15{{\sec }^{3}}x{{\tan }^{2}}x+3{{\sec }^{3}}x{{\tan }^{2}}x \\
& +\sec x{{\tan }^{4}}x \\
& {{f}^{4}}\left( x \right)=5{{\sec }^{5}}x+18{{\sec }^{3}}x{{\tan }^{2}}x+\sec x{{\tan }^{4}}x \\
& \\
& \text{Calculate }{{f}^{n}}\left( c \right)\text{ at }c=0 \\
& f\left( 0 \right)=\sec \left( 0 \right)=1 \\
& f'\left( 0 \right)=\sec \left( 0 \right)\tan \left( 0 \right)=0 \\
& f''\left( 0 \right)={{\sec }^{3}}\left( 0 \right)+\sec \left( 0 \right){{\tan }^{2}}\left( 0 \right)=1 \\
& f'''\left( 0 \right)=5{{\sec }^{3}}\left( 0 \right)\tan \left( 0 \right)+\sec \left( 0 \right){{\tan }^{3}}\left( 0 \right)=0 \\
& {{f}^{4}}\left( 0 \right)=5{{\sec }^{5}}\left( 0 \right)+18{{\sec }^{3}}\left( 0 \right){{\tan }^{2}}\left( 0 \right)+\sec \left( 0 \right){{\tan }^{4}}\left( 0 \right) \\
& {{f}^{4}}\left( 0 \right)=5\left( 1 \right)+18\left( 1 \right)\left( 0 \right)+\left( 1 \right)\left( 0 \right)=5 \\
& \\
& \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 \\
& \sec \left( x \right)=\frac{{{f}^{\left( 0 \right)}}\left( 0 \right)}{0!}\left( x-0 \right)+\frac{f'\left( 0 \right)}{1!}\left( x-1 \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} \\
& \sec \left( x \right)=\frac{1}{1}{{\left( x \right)}^{0}}+\frac{0}{1}\left( x \right)+\frac{1}{2!}{{\left( x \right)}^{2}}+\frac{0}{6!}{{\left( x \right)}^{3}}+\frac{5}{4!}{{\left( x \right)}^{4}}+\cdots \\
& \sec \left( x \right)=1+\frac{1}{2!}{{x}^{2}}+\frac{5}{4!}{{x}^{4}}+\cdots \\
\end{align}\]
