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
$${we}\:{have}\:\:\:{forf}\left({t}\right)={arctant} \\ $$$${f}^{'} \left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:{with}\:−\mathrm{1}<{t}<\mathrm{1}\Rightarrow{f}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} \\ $$$$\Rightarrow{f}\left({t}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow{for}\:−\frac{\pi}{\mathrm{4}}<\theta<\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({tan}\theta\right)=\:\theta\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{\left({tan}\theta\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$={tan}\theta\:−\frac{\mathrm{1}}{\mathrm{3}}\left({tan}\theta\right)^{\mathrm{3}} \:\:+\frac{\mathrm{1}}{\mathrm{5}}\:\left({tan}\theta\right)^{\mathrm{5}} \:+.... \\ $$