Question Number 201561 by Mingma last updated on 09/Dec/23
Answered by ajfour last updated on 11/Dec/23
Commented by ajfour last updated on 11/Dec/23
$$\mathrm{2}\theta+\mathrm{2}\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\left(\theta+\phi\right)=\mathrm{1} \\ $$$$\mathrm{tan}\:\theta+\mathrm{tan}\:\phi=\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi\:\:\:\: \\ $$$$\mathrm{tan}\:\theta=\frac{{q}}{{r}}=\frac{{r}}{{p}+{a}}\:\Rightarrow\:\:{pq}+{aq}={r}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\phi=\frac{{p}}{{r}}=\frac{{r}}{{q}+{b}}\:\Rightarrow\:{pq}+{bp}={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:\phi}=\frac{{q}}{{p}}\:\:\:\&\:\:{aq}={bp}\:\:\left({subtracting}\right) \\ $$$${hence}\:\:\:\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:\phi}=\frac{{b}}{{a}} \\ $$$$\mathrm{2}{pq}+{aq}+{bp}=\mathrm{2}{r}^{\mathrm{2}} \:\:\:\:\:\left({adding}\right) \\ $$$$\mathrm{2}\left({r}\mathrm{tan}\:\phi\right)\left({r}\mathrm{tan}\:\theta\right)+{r}\left({a}\mathrm{tan}\:\theta+{b}\mathrm{tan}\:\phi\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{r}}\left({a}\mathrm{tan}\:\theta+{b}\mathrm{tan}\:\phi\right)=\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi \\ $$$${but}\:\:\:\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi=\mathrm{tan}\:\theta+\mathrm{tan}\:\phi \\ $$$$\&\:\:{b}\mathrm{tan}\:\phi={a}\mathrm{tan}\:\theta\:\:\:\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{r}}\left(\mathrm{2}{a}\mathrm{tan}\:\theta\right)=\mathrm{tan}\:\theta+\frac{{a}}{{b}}\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}} \\ $$