$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$

$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_1}{n_2}$

$n_1 \cos \theta_1 = n_2 \cos \theta_2$

$n_1 \tan \theta_1 = n_2 \tan \theta_2$