Answer
$0$
Work Step by Step
$60^o$ is a special angle whose sine value is $\dfrac{\sqrt3}{2}$.
$30^o$ is a special angle whose cosine value is $\dfrac{\sqrt3}{2}.$
Thus,
$\sin^2{60^o} - \cos^2{30^o}
\\= \left(\dfrac{\sqrt3}{2}\right)^2 - \left(\dfrac{\sqrt3}{2}\right)^2
\\=0$