Question Number 221406 by ajfour last updated on 03/Jun/25
Commented by ajfour last updated on 03/Jun/25
$${Find}\:\mathrm{tan}\:\phi\:\:\:{if}\:\mathrm{tan}\:\theta=\mathrm{1}/\mathrm{4}. \\ $$
Answered by mr W last updated on 03/Jun/25
$$\mathrm{tan}\:\theta=\frac{\theta−\theta\:\mathrm{sin}\:\phi}{\theta+\theta\:\mathrm{cos}\:\phi} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{1}−\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\frac{\mathrm{1}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}=\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${t}=\mathrm{1}−\sqrt{\mathrm{2}\:\mathrm{tan}\:\theta}=\mathrm{tan}\:\frac{\phi}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}−\sqrt{\mathrm{2}\:\mathrm{tan}\:\theta}}{\:\sqrt{\mathrm{tan}\:\theta}\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{tan}\:\theta}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}}\approx\mathrm{0}.\mathrm{641} \\ $$
Commented by ajfour last updated on 03/Jun/25
$${Excellent}\:{sir}.\:{concise}\:{and}\:{correct}! \\ $$