Answer
$ \dfrac{1}{2}$
Work Step by Step
Let us consider:
$g(\theta)=\cos \theta$ and $\theta=60^{\circ}$
We know from the unit circle that
$\cos (60^{\circ})=\cos (\dfrac{\pi}{3})=\dfrac{1}{2}$
Since, $\cos$, $\sec$ are even trigonometric functions, it follows that $f (-x)=f(x) \implies \cos(-x)=\cos(x)$
Thus, we have: $g(-\theta)= \cos (-60^{\circ})
\\= \cos 60^{\circ} \\= \cos (\dfrac{\pi}{3}) \\ = \dfrac{1}{2}$