If two languages together cover all words and one is regular, is the other one as well?
In fact, the answer is no. 

If $L_1$$\subseteq$ $\Sigma^*$, $L_2$$\subseteq$ $\Sigma^*$ , $L_1$ is regular and $L_1$$\cup$ $L_2$ = $\Sigma^*$ then is $L_2$ is not necessarily regular.

We can prove this through counter-example.

If we let $L_1$ = $\Sigma^*$, then we can choose any non-regular language in $\Sigma$  for $L_2$.

If we take $\Sigma$ = {a,b} and then let $L_2$ = $a^n$$b^n$ (A non-regular language) then  $L_1$$\cup$ $L_2$ = $\Sigma^*$ and $L_2$ is non-regular as required.