Menu Close

Question-201561




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
2θ+2φ=(π/2)  tan (θ+φ)=1  tan θ+tan φ=1−tan θtan φ      tan θ=(q/r)=(r/(p+a)) ⇒  pq+aq=r^2   tan φ=(p/r)=(r/(q+b)) ⇒ pq+bp=r^2   ⇒  ((tan θ)/(tan φ))=(q/p)   &  aq=bp  (subtracting)  hence   ((tan θ)/(tan φ))=(b/a)  2pq+aq+bp=2r^2      (adding)  2(rtan φ)(rtan θ)+r(atan θ+btan φ)         =2r^2   ⇒ (1/(2r))(atan θ+btan φ)=1−tan θtan φ  but   1−tan θtan φ=tan θ+tan φ  &  btan φ=atan θ    ⇒  (1/(2r))(2atan θ)=tan θ+(a/b)tan θ  ⇒   (1/r)=(1/a)+(1/b)
$$\mathrm{2}\theta+\mathrm{2}\phi=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\left(\theta+\phi\right)=\mathrm{1} \\ $$$$\mathrm{tan}\:\theta+\mathrm{tan}\:\phi=\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi\:\:\:\: \\ $$$$\mathrm{tan}\:\theta=\frac{{q}}{{r}}=\frac{{r}}{{p}+{a}}\:\Rightarrow\:\:{pq}+{aq}={r}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\phi=\frac{{p}}{{r}}=\frac{{r}}{{q}+{b}}\:\Rightarrow\:{pq}+{bp}={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:\phi}=\frac{{q}}{{p}}\:\:\:\&\:\:{aq}={bp}\:\:\left({subtracting}\right) \\ $$$${hence}\:\:\:\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:\phi}=\frac{{b}}{{a}} \\ $$$$\mathrm{2}{pq}+{aq}+{bp}=\mathrm{2}{r}^{\mathrm{2}} \:\:\:\:\:\left({adding}\right) \\ $$$$\mathrm{2}\left({r}\mathrm{tan}\:\phi\right)\left({r}\mathrm{tan}\:\theta\right)+{r}\left({a}\mathrm{tan}\:\theta+{b}\mathrm{tan}\:\phi\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{r}}\left({a}\mathrm{tan}\:\theta+{b}\mathrm{tan}\:\phi\right)=\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi \\ $$$${but}\:\:\:\mathrm{1}−\mathrm{tan}\:\theta\mathrm{tan}\:\phi=\mathrm{tan}\:\theta+\mathrm{tan}\:\phi \\ $$$$\&\:\:{b}\mathrm{tan}\:\phi={a}\mathrm{tan}\:\theta\:\:\:\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{r}}\left(\mathrm{2}{a}\mathrm{tan}\:\theta\right)=\mathrm{tan}\:\theta+\frac{{a}}{{b}}\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *