Question Number 109303 by peter frank last updated on 22/Aug/20
Commented by som(math1967) last updated on 23/Aug/20
$$\mathrm{Acute}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{y}=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:\mathrm{istan}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore\mathrm{angle}\:\mathrm{between}\:\mathrm{bisector} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:=\frac{\pi}{\mathrm{8}} \\ $$$$\therefore\mathrm{gradiant}\:\mathrm{of}\:\mathrm{angle}\:\mathrm{bisector} \\ $$$$=\mathrm{tan}\frac{\pi}{\mathrm{8}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$