$$ (sin60^o)(cos30^o)(sin30^o)(cos60^o)=? $$

As we know that

$$ sin60^o=\frac{\sqrt3}{2},\quad cos30^o=\frac{\sqrt3}{2},\quad sin30^o=\frac{1}{2},\quad cos60^o=\frac{1}{2} $$

So, let's plug the values into the given equation

$$ (sin60^o)(cos30^o)(sin30^o)(cos60^o)=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) $$

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

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

$$ =1 $$

Also, we can see that

$$ sin(60^o+30^o)=sin90^o=1 $$