Question Number 74966 by ajfour last updated on 05/Dec/19
$${If}\:\:\mathrm{sin}\:\mathrm{2}{A}=\lambda\mathrm{sin}\:\mathrm{2}{B} \\ $$$${Prove}\:{that}\:\:\frac{\mathrm{tan}\:\left({A}+{B}\right)}{\mathrm{tan}\:\left({A}−{B}\right)}=\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}}\:. \\ $$
Commented by Prithwish sen last updated on 05/Dec/19
$$\frac{\mathrm{sin2A}}{\mathrm{sin2B}}\:=\:\lambda \\ $$$$\frac{\mathrm{sin2A}+\mathrm{sin2B}}{\mathrm{sin2A}−\mathrm{sin2B}}\:=\:\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}} \\ $$$$\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)\mathrm{sin}\left(\mathrm{A}−\mathrm{B}\right)}\:=\:\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}} \\ $$$$\frac{\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}\right)}{\boldsymbol{\mathrm{tam}}\left(\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}\right)}\:=\:\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\lambda}+\mathrm{1}}\:\:\boldsymbol{\mathrm{proved}}. \\ $$
Commented by ajfour last updated on 05/Dec/19
$${Thanks}. \\ $$
Answered by ajfour last updated on 05/Dec/19
![sin [(A+B)+(A−B)]=λsin [A+B−(A−B)] ⇒ sin (A+B)cos (A−B)+cos (A+B)sin (A−B) −λ{sin (A+B)cos (A−B)−cos (A+B)sin (A−B)=0 dividing by cos (A+B)cos (A−B) tan (A+B)+tan (A−B) = λ{tan (A+B)−tan (A−B) ⇒ ((tan (A+B))/(tan (A−B)))=((λ+1)/(λ−1)) .](https://www.tinkutara.com/question/Q74972.png)
$$\mathrm{sin}\:\left[\left({A}+{B}\right)+\left({A}−{B}\right)\right]=\lambda\mathrm{sin}\:\left[{A}+{B}−\left({A}−{B}\right)\right] \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{cos}\:\left({A}+{B}\right)\mathrm{sin}\:\left({A}−{B}\right) \\ $$$$−\lambda\left\{\mathrm{sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)−\mathrm{cos}\:\left({A}+{B}\right)\mathrm{sin}\:\left({A}−{B}\right)=\mathrm{0}\right. \\ $$$${dividing}\:{by}\:\mathrm{cos}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right) \\ $$$$\mathrm{tan}\:\left({A}+{B}\right)+\mathrm{tan}\:\left({A}−{B}\right) \\ $$$$\:\:\:\:\:=\:\lambda\left\{\mathrm{tan}\:\left({A}+{B}\right)−\mathrm{tan}\:\left({A}−{B}\right)\right. \\ $$$$\Rightarrow\:\:\frac{\mathrm{tan}\:\left({A}+{B}\right)}{\mathrm{tan}\:\left({A}−{B}\right)}=\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}}\:. \\ $$
Commented by Prithwish sen last updated on 05/Dec/19
$$\mathrm{Wow}\:! \\ $$