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Question Number 143390 by ZiYangLee last updated on 13/Jun/21

Given that (((tan α)/(sin θ))−((tan β)/(tan θ)))^2 =tan^2 α−tan^2 β, prove that cos θ=((tan β)/(tan α ))

$$\mathrm{Given}\:\mathrm{that}\:\left(\frac{\mathrm{tan}\:\alpha}{\mathrm{sin}\:\theta}−\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} =\mathrm{tan}^{\mathrm{2}} \alpha−\mathrm{tan}^{\mathrm{2}} \beta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\:\theta=\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\alpha\:} \\ $$

Answered by mindispower last updated on 13/Jun/21

⇔((1/(sin(θ)))−(1/(tg(θ))).((tg(β))/(tg(α))))^2 =1−((tg^2 (β))/(tg^2 (α))) let X=((tg(β))/(tg(α))) ((1/(sin(θ)))−(X/(tg(θ))))^2 =1−X^2 ⇔X^2 (1+(1/(tg^2 (θ))))−2((cos(θ))/(sin^2 (θ)))X+((cos^2 (θ))/(sin^2 (θ)))=0 X^2 −2cos(θ)X+cos^2 (θ)=0 (X−cos(θ))^2 =0⇒X=cos(θ)

$$\Leftrightarrow\left(\frac{\mathrm{1}}{{sin}\left(\theta\right)}−\frac{\mathrm{1}}{{tg}\left(\theta\right)}.\frac{{tg}\left(\beta\right)}{{tg}\left(\alpha\right)}\right)^{\mathrm{2}} =\mathrm{1}−\frac{{tg}^{\mathrm{2}} \left(\beta\right)}{{tg}^{\mathrm{2}} \left(\alpha\right)} \\ $$$${let}\:{X}=\frac{{tg}\left(\beta\right)}{{tg}\left(\alpha\right)} \\ $$$$\left(\frac{\mathrm{1}}{{sin}\left(\theta\right)}−\frac{{X}}{{tg}\left(\theta\right)}\right)^{\mathrm{2}} =\mathrm{1}−{X}^{\mathrm{2}} \\ $$$$\Leftrightarrow{X}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{tg}^{\mathrm{2}} \left(\theta\right)}\right)−\mathrm{2}\frac{{cos}\left(\theta\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}{X}+\frac{{cos}^{\mathrm{2}} \left(\theta\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}=\mathrm{0} \\ $$$${X}^{\mathrm{2}} −\mathrm{2}{cos}\left(\theta\right){X}+{cos}^{\mathrm{2}} \left(\theta\right)=\mathrm{0} \\ $$$$\left({X}−{cos}\left(\theta\right)\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{X}={cos}\left(\theta\right) \\ $$$$ \\ $$

Commented by ZiYangLee last updated on 14/Jun/21

thank you.

$${thank}\:{you}. \\ $$

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