Given: $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))$
We know that $\cos^{-1} (-x) = \pi - \cos^{-1}x$
$\Rightarrow$ $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))\; $$= \cos \large ( \frac{\pi}{2} -$$ \pi + \cos^{-1} \large (\frac{\sqrt 3}{2}))$ = $\cos \large ( $$\cos^{-1} \large (\frac{\sqrt 3}{2}) - \frac{\pi}{2})$
We know that $\cos^{-1} \large (\frac{\sqrt 3}{2}) $$= \large \frac{\pi}{6}$
$\Rightarrow$ $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))\; $ $= \cos \large (\frac{\pi}{6} - \frac{\pi}{2})$$ = \cos - \large (\frac{\pi}{3})$
We know that: $\cos - \large (\frac{\pi}{3}) $$= \frac{1}{2}$
$\Rightarrow \cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))$ $= \large \frac{1}{2}$