Question Number 197524 by sonukgindia last updated on 20/Sep/23
Answered by som(math1967) last updated on 20/Sep/23
$$\mathrm{2arc}{tan}\left(\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{arc}{tan}\left\{\frac{\frac{\mathrm{2}{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }}\right\} \\ $$$$=\mathrm{arc}{tan}\left\{\frac{\mathrm{2}{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}×\frac{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right\} \\ $$$$=\mathrm{arc}{tan}\left(\frac{\mathrm{2}{x}}{\mathrm{2}}\right)=\mathrm{arc}{tanx} \\ $$
Answered by Mathspace last updated on 20/Sep/23
$${x}={tant}\:\Rightarrow \\ $$$$\mathrm{2}{arctan}\left(\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{2}{artan}\left(\frac{{tant}}{\mathrm{1}+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}\right) \\ $$$$=\mathrm{2}{arctan}\left(\frac{{tant}}{\mathrm{1}+\frac{\mathrm{1}}{{cost}}}\right) \\ $$$$=\mathrm{2}{arctan}\left(\frac{{cost}\:{tant}}{\mathrm{1}+{cost}}\right) \\ $$$$=\mathrm{2}{arctan}\left(\frac{{sint}}{\mathrm{1}+{cost}}\right) \\ $$$$=\mathrm{2}{arctan}\left(\frac{\mathrm{2}{sin}\left(\frac{{t}}{\mathrm{2}}\right){cos}\left(\frac{{t}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)}\right) \\ $$$$=\mathrm{2}{arctan}\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{2}.\frac{{t}}{\mathrm{2}}={t}={arctanx} \\ $$