When unpolarized light is incident from air onto a transparent...

According to Brewster's law, the refractive index ($\mu$) of the medium is related to Brewster's angle ($i_p$) by:
$\mu = \tan(i_p)$

For total internal reflection at the medium-air interface, the critical angle ($\theta_c$) is related to the refractive index by:
$\sin(\theta_c) = \dfrac{1}{\mu}$

Substituting the value of $\mu$ from Brewster's law into the critical angle formula, we get:
$\sin(\theta_c) = \dfrac{1}{\tan(i_p)}$

$\sin(\theta_c) = \cot(i_p)$

Answer: $\sin(\theta_c) = \cot(i_p)$