Find the angle θ

Given $\cot \theta = \frac{1}{\sqrt{3}}$. Since $\cot 60° = \frac{1}{\sqrt{3}}$, we find $\theta = 60°$.

$\theta = 60°$

Evaluate the first part of the expression

Substitute $\theta = 60°$ into $\frac{2 - \sin^{2} \theta}{1 + \cos^{2} \theta}$. With $\sin 60° = \frac{\sqrt{3}}{2}$ and $\cos 60° = \frac{1}{2}$, we get $\frac{2 - (\frac{3}{4})}{1 + (\frac{1}{4})} = \frac{\frac{5}{4}}{\frac{5}{4}} = 1$.

$\frac{2 - \frac{3}{4}}{1 + \frac{1}{4}} = 1$

Evaluate the second part of the expression

Calculate $\csc^{2} 60° + \sec^{2} 60°$. Since $\csc 60° = \frac{2}{\sqrt{3}}$ and $\sec 60° = 2$, the value is $\frac{4}{3} + 4 = \frac{16}{3}$.

$\csc^{2} 60° + \sec^{2} 60° = \frac{16}{3}$

Final Sum

Add the two results: $1 + \frac{16}{3} = \frac{3 + 16}{3} = \frac{19}{3}$.

$1 + \frac{16}{3} = \frac{19}{3}$