If ${\left(\frac{\sin \theta }{\sin \varphi }\right)}^2=\frac{\tan \theta }{\tan \varphi }=3$ , then

Given $\frac{{\sin }^2\theta }{{\sin }^2\varphi }=3$

${\sin }^2\theta =3{\sin }^2\varphi$

$1-\cos 2\theta =3-3\cos 2\varphi$

$3{\cos }^2\varphi -2={\cos }^2\theta$

$\Rightarrow \frac{3-2{\sec }^2\varphi }{{\sec }^2\varphi }=\frac{1}{{\sec }^2\theta }$

$\Rightarrow {\sec }^2\theta =\frac{{\sec }^2\varphi }{3-2{\sec }^2\varphi }\to \left(1\right)$

$\Rightarrow \frac{\tan \theta }{\tan \varphi }=3$

$\Rightarrow \tan \theta =3\tan \varphi$

$\Rightarrow {\tan }^2\theta =9{\tan }^2\varphi \to \left(2\right)$

$\Rightarrow {\sec }^2\theta -1=9{\tan }^2\varphi$

$\Rightarrow \frac{{\sec }^2\varphi }{3-2{\sec }^2\varphi }-1=9{\tan }^2\varphi \left(\text{from}\left(1\right)\right)$

$\Rightarrow \frac{{\sec }^2\varphi -3+2{\sec }^2\varphi }{3-2{\sec }^2\varphi }=9{\tan }^2\varphi$

$\Rightarrow \frac{3\left[{\sec }^2\varphi -1\right]}{3-2{\sec }^2\varphi }=9{\tan }^2\varphi$

$\Rightarrow {\tan }^2\varphi =3{\tan }^2\varphi \left[3-2{\sec }^2\varphi \right]$

$\Rightarrow 1=9-6{\sec }^2\varphi$

$\Rightarrow 1=3-6\left[{\sec }^2\varphi -1\right]$

$\Rightarrow 6{\tan }^2\varphi =2$

$\Rightarrow {\tan }^2\varphi =\frac{1}{3}$

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

From (2)

$\Rightarrow {\tan }^2\theta =9{\tan }^2\varphi$

$=9\times \frac{1}{3}=3$

$\therefore \tan \theta =\pm \sqrt{3}$