If Brewster's angle be 'θ', then the critical angle is

$\sin^{-1} (\cot θ)$

The correct answer is Option (1) → $\sin^{-1} (\cot θ)$

According to Brewster's Angle,

$\tan θ_B=\frac{μ_1}{μ_2}$   ...(1)

and, Critical Angle is,

$\sin θ_C=\frac{μ_1}{μ_2}$   ...(2)

According to (1) and (2),

$\sin θ_C=\frac{1}{\tan θ_B}=\cot θ_B$

$∴θ_C=\sin^{-1} (\cot θ_B)$