Answer: Using Taylor's series, we know that any function can be expanded in the form near the point $x=x_0.$

$f(x)=f(x_0)+(x-x_0)\dfrac{f'(x_0)}{1!}+(x-x_0)^2 \dfrac{f''(x_0)}{2!}+(x-x_0)^3 \dfrac{f'''(x_0)}{3!}+(x-x_0)^4 \dfrac{f^{iv}(x_0)}{4!}+...$

Here $x_0=0.$ Thus, 

$f(x)=f(0)+xf'(0)+x^2 \dfrac{f''(0)}{2!}+x^3 \dfrac{f'''(0)}{3!}+x^4 \dfrac{f^{iv}(0)}{4!}+...$............................(1)

Here, $f(x)=sec^2x, f'(x)=2sec^2xtanx, f''(x)=2[sec^4x+2sec^2xtan^2x] $

$f'''(x)=2[4sec^4xtanx+4sec^2xtan^3x+4sec^4xtanx]$

$f^{iv}(x)=2[4sec^6x+16sec^4xtan^2x+8sec^2xtan^4x+12sec^4xtan^2x+4sec^6x+16sec^4xtan^2x]$

Also, $f(0)=1, f'(0)=0, f''(0)=2, f'''(0)=0, f^{iv}(0)=16.$

Thus substituting it into (1), we get

$f(x)=1+0+x^2 \dfrac{2}{2!}+ \dfrac{x^3}{3!}(0)+\dfrac{x^4}{4!} (16)+...=1+x^2+\dfrac{2}{3}x^4+...$                          Hence Proved.