Question

Solve ${\left(\frac{\sin \theta }{\sin \varphi }\right)}^2=\frac{\tan \theta }{\tan \varphi }=3$.

Answer 1

The given equation is ${\left(\frac{\sin \theta }{\sin \varphi }\right)}^2=\frac{\sin \theta }{\sin \varphi }\cdot \frac{\cos \varphi }{\cos \theta }$

or $\frac{\sin \theta }{\sin \varphi }=\frac{\cos \varphi }{\cos \theta }$

[Note $\frac{sin\theta }{sin\varphi }\ne 0$, since its square is given to be $3$]

or $2\sin \theta \cos \theta =2\sin \varphi \cos \varphi$

$\Rightarrow \sin 2\theta =\sin 2\varphi$

$\Rightarrow \frac{2\tan \theta }{(1+{\tan }^2\theta )}=\frac{2\tan \varphi }{(1+{\tan }^2\varphi )}$

$\Rightarrow \frac{6\tan \varphi }{(1+9{\tan }^2\varphi )}=\frac{2\tan \varphi }{(1+{\tan }^2\varphi )}$

$[\because \tan \theta =3\tan \varphi ]$

$\Rightarrow \frac{3}{(1+9{\tan }^2\varphi )}=\frac{1}{(1+{\tan }^2\varphi )}$ $[\because \tan \varphi \ne 0]$

$\Rightarrow 3+3{\tan }^2\varphi =1+9{\tan }^2\varphi$ $\Rightarrow 6{\tan }^2\varphi =2$

$\Rightarrow \tan \varphi =\pm 1/\sqrt{3}$

$\therefore \varphi =n\pi \pm \left(\pi /6\right)$

Now $\tan \theta =3\tan \varphi =3(\pm 1/\sqrt{3})=\pm \sqrt{3}$

$\therefore \theta =n\pi \pm \left(\pi /3\right)$.