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Question Number 34115 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

Commented by abdo mathsup 649 cc last updated on 03/May/18

$$wehaveforf\left(t\right)=arctant$$$$f^{{}^{\prime }}\left(t\right)=\frac{1}{1+t^2}with-1<t<1\Rightarrow f^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}$$$$\Rightarrow f\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{t^{2n+1}}{2n+1}\Rightarrow for-\frac{\pi }{4}<\theta <\frac{\pi }{4}$$$$f\left(tan\theta \right)=\theta =\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{{\left(tan\theta \right)}^{2n+1}}{2n+1}$$$$=tan\theta -\frac{1}{3}{\left(tan\theta \right)}^3+\frac{1}{5}{\left(tan\theta \right)}^5+....$$

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