Question

Solve for $\theta$ and $\varphi$ the equation $\sec \theta =\sqrt{2}\sec \varphi ,\tan \theta =\sqrt{3}\tan \varphi$.

Answer 1

Eliminating $\theta$, we get $2se{c}^2\varphi -3{\tan }^2\varphi =1$
$\therefore {\tan }^2\varphi =1={\tan }^2\frac{\pi }{4}$ or $\varphi =n\pi \pm \frac{\pi }{4}$
Eliminating $\varphi$, we get ${\tan }^2\theta =3={\tan }^2\frac{\pi }{3}$
$\therefore \theta =n\pi \pm \frac{\pi }{3}$.