Question Number 206421 by MATHEMATICSAM last updated on 13/Apr/24
$$\mathrm{If}\:\mathrm{tan}{p}\theta\:=\:{p}\mathrm{tan}\theta\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}^{\mathrm{2}} {p}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\:\frac{{p}^{\mathrm{2}} }{\mathrm{1}\:+\:\left({p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \theta}\:.\: \\ $$
Answered by Frix last updated on 13/Apr/24
$$\mathrm{tan}\:{p}\theta\:={p}\mathrm{tan}\:\theta\:={t} \\ $$$$\Leftrightarrow \\ $$$${p}\theta=\mathrm{tan}^{−\mathrm{1}} \:{t}\:\wedge\theta=\mathrm{tan}^{−\mathrm{1}} \:\frac{{t}}{{p}} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{p}\theta\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\wedge\mathrm{sin}^{\mathrm{2}} \:\theta\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +{p}^{\mathrm{2}} } \\ $$$$\frac{\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}}{\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +{p}^{\mathrm{2}} }}=\frac{{p}^{\mathrm{2}} }{\mathrm{1}+\left({p}^{\mathrm{2}} −\mathrm{1}\right)\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +{p}^{\mathrm{2}} }}\:\mathrm{which}\:\mathrm{is}\:\mathrm{true} \\ $$$$ \\ $$