f($x) \;=\; tan^{-1} x \therefore f(1) \;=\; tan^{-1} 1 \;=\; \dfrac{\pi}{4} $

Differentiating w.r.t.x,

$ f_1 (x)= \dfrac{1}{1+x^2 } \; \; \; \therefore f_1 (1)= \dfrac{1}{1+1^2 } = \dfrac{1}{2 } $

Differentiating again w.r.t.x,

$ \therefore f_2 (x)= \dfrac{ -2x}{(1+x^2 )^2} \; \; \; \therefore f_2 (1)= \dfrac{-2}{(1+1)^2} = \dfrac{-1}{2 } $

$ f_3 (x)= \dfrac{(1+x^2 )^2 (-2)+2x.2(1+x^2 )(2x)}{(1+x^2 )^4} \\ \; \\ \; \\ \therefore f_3 (1)= \dfrac{(1+1^2 )^2 (-2)+2.2(1+1^2 )(2)} {(1+1^2 )^4} \; = \; \dfrac{-8+16}{16}= \dfrac{1}{2} $

By using Taylor' series,

$ f(x)=f(a)+(x-a) f'(a)+ \dfrac{(x-a)^2}{2!} f'' (a)+ \dfrac{(x-a)^3}{3!} f''' (a) + \ldots \\ \; \\ \; \\ \therefore tan^{-1}⁡x = \dfrac{\pi}{4} +\dfrac{1}{2} (x-1) - \dfrac{1}{2} \dfrac{(x-1)^2}{2!} +\dfrac{1}{2} \dfrac{(x-1)^3}{3!} + \ldots \\ \; \\ \; \\ \therefore tan^{-1}⁡x = \dfrac{\pi}{4} +\dfrac{1}{2} (x-1) - \dfrac{(x-1)^2}{4} + \dfrac{(x-1)^3}{12} + \ldots $