Question Number 18695 by Arnab Maiti last updated on 27/Jul/17
$$\mathrm{if}\:\:\frac{\mathrm{n}\:\mathrm{tan}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\theta\right)}=\frac{\mathrm{m}\:\mathrm{tan}\left(\alpha−\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}\theta=\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\mathrm{tan}\alpha\right) \\ $$
Answered by Tinkutara last updated on 28/Jul/17
$${n}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:=\:{m}\:\mathrm{sin}\:\left(\alpha\:−\:\theta\right)\:\mathrm{cos}\:\left(\alpha\:−\:\theta\right) \\ $$$$\frac{{n}}{{m}}\:=\:\frac{\mathrm{sin}\:\left(\alpha\:−\:\theta\right)\:\mathrm{cos}\:\left(\alpha\:−\:\theta\right)}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}\:=\:\frac{\mathrm{sin}\:\mathrm{2}\left(\alpha\:−\:\theta\right)}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\frac{{n}\:−\:{m}}{{n}\:+\:{m}}\:=\:\frac{\mathrm{2}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\left(\alpha\:−\:\mathrm{2}\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\left(\alpha\:−\:\mathrm{2}\theta\right)} \\ $$$$\frac{{n}\:−\:{m}}{{n}\:+\:{m}}\:\mathrm{tan}\:\alpha\:=\:\mathrm{tan}\:\left(\alpha\:−\:\mathrm{2}\theta\right) \\ $$$$\alpha\:=\:\mathrm{2}\theta\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}\:−\:{m}}{{n}\:+\:{m}}\:\mathrm{tan}\:\alpha\right) \\ $$
Commented by Arnab Maiti last updated on 28/Jul/17
$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{very}\:\mathrm{much}.\:\mathrm{I}\:\mathrm{really} \\ $$$$\mathrm{appriciate}\:\mathrm{you}\:\mathrm{sir}!! \\ $$$$ \\ $$