Question Number 143390 by ZiYangLee last updated on 13/Jun/21
$$\mathrm{Given}\:\mathrm{that}\:\left(\frac{\mathrm{tan}\:\alpha}{\mathrm{sin}\:\theta}−\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} =\mathrm{tan}^{\mathrm{2}} \alpha−\mathrm{tan}^{\mathrm{2}} \beta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\:\theta=\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\alpha\:} \\ $$
Answered by mindispower last updated on 13/Jun/21
$$\Leftrightarrow\left(\frac{\mathrm{1}}{{sin}\left(\theta\right)}−\frac{\mathrm{1}}{{tg}\left(\theta\right)}.\frac{{tg}\left(\beta\right)}{{tg}\left(\alpha\right)}\right)^{\mathrm{2}} =\mathrm{1}−\frac{{tg}^{\mathrm{2}} \left(\beta\right)}{{tg}^{\mathrm{2}} \left(\alpha\right)} \\ $$$${let}\:{X}=\frac{{tg}\left(\beta\right)}{{tg}\left(\alpha\right)} \\ $$$$\left(\frac{\mathrm{1}}{{sin}\left(\theta\right)}−\frac{{X}}{{tg}\left(\theta\right)}\right)^{\mathrm{2}} =\mathrm{1}−{X}^{\mathrm{2}} \\ $$$$\Leftrightarrow{X}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{tg}^{\mathrm{2}} \left(\theta\right)}\right)−\mathrm{2}\frac{{cos}\left(\theta\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}{X}+\frac{{cos}^{\mathrm{2}} \left(\theta\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}=\mathrm{0} \\ $$$${X}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){X}+{cos}^{\mathrm{2}} \left(\theta\right)=\mathrm{0} \\ $$$$\left({X}−{cos}\left(\theta\right)\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{X}={cos}\left(\theta\right) \\ $$$$ \\ $$
Commented by ZiYangLee last updated on 14/Jun/21
$${thank}\:{you}. \\ $$