Let

$f(x)=tan^{-1}x$  and $a=1 $

therefore, $ f^{'}(x)={1\over1+x^2}$.....(i)

$f^{''}(x)=-2x.({1+{x^2}})^{-2}$....(ii)

$f^{'''}(x)=-2({1+x^2})^{-2} + 4x.2x({1+x^2})^{-3}$....(iii)

At a=1,

$f(a)=pi/4$

$f^{'}(a)=1/2$

$f^{''}(a)=-1/2$

$f^{'''}(a)=1/2$

By Taylor Series,

$f(x)=f(a)+(x-a)f^{'}(a)+{(x-a)^{2}}f^{''}(a)/2!+(x-a)^{3}f^{'''}(a)/3!+...$

therefore,$f(x)=pi/4+(x-1)/2-(x-1)^{2}/2.2!+(x-1)^{3}/2.3!+...$

therefore,

$tan^{-1}x=pi/4+(x-1)/2-(x-1)^{2}/4+(x-1)^{3}/12+...$