Question Number 225241 by Spillover last updated on 18/Oct/25
Answered by som(math1967) last updated on 19/Oct/25
$$\:{tan}\theta=\frac{{nsin}\alpha{cos}\alpha.{sec}^{\mathrm{2}} \alpha}{\left(\mathrm{1}−{nsin}^{\mathrm{2}} \alpha\right){sec}^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow{tan}\theta=\frac{{ntan}\alpha}{{sec}^{\mathrm{2}} \alpha−{ntan}^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{ntan}\alpha}{\mathrm{1}+\left(\mathrm{1}−{n}\right){tan}^{\mathrm{2}} \alpha} \\ $$$$\:{tan}\left(\theta−\alpha\right)=\frac{{tan}\theta−{tan}\alpha}{\mathrm{1}+{tan}\theta{tan}\alpha} \\ $$$$=\frac{\frac{{ntan}\alpha}{\mathrm{1}+\left(\mathrm{1}−{n}\right){tan}^{\mathrm{2}} \alpha}−{tan}\alpha}{\mathrm{1}+\frac{{ntan}^{\mathrm{2}} \alpha}{\mathrm{1}+\left(\mathrm{1}−{n}\right){tan}^{\mathrm{2}} \alpha}} \\ $$$$=\frac{{tan}\alpha\left\{\frac{{n}−\mathrm{1}−{tan}^{\mathrm{2}} \alpha+{ntan}^{\mathrm{2}} \alpha}{\mathrm{1}+\left(\mathrm{1}−{n}\right){tan}^{\mathrm{2}} \alpha}\right\}}{\frac{\mathrm{1}+{tan}^{\mathrm{2}} \alpha−{ntan}^{\mathrm{2}} \alpha+{ntan}^{\mathrm{2}} \alpha}{\mathrm{1}+\left(\mathrm{1}−{n}\right){tan}^{\mathrm{2}} \alpha}} \\ $$$$=\frac{\left\{{n}\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)−\mathrm{1}\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\right\}{tan}\alpha}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)} \\ $$$$=\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\left({n}−\mathrm{1}\right){tan}\alpha}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)} \\ $$$$=\left({n}−\mathrm{1}\right){tan}\alpha \\ $$
Commented by Spillover last updated on 25/Nov/25
$${thanks} \\ $$