Question Number 113823 by deepraj123 last updated on 15/Sep/20 | ||
$$\mathrm{A}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\:=\:\frac{\pi}{\mathrm{2}}\:\:\mathrm{is} \\ $$ | ||
Answered by MJS_new last updated on 15/Sep/20 | ||
$${x}=\mathrm{0}\:\mathrm{because}\:\mathrm{arctan}\:\mathrm{1}\:=\frac{\pi}{\mathrm{4}} \\ $$ | ||