@Pratik, I also think the same. It should be minimum vertex cover and largest independent set.
Because if we consider 2 theorems , then answer will be valid in case of minimum vertex cover not vertex cover.
1) size of largest independent set + size of minimum vertex cover = no. of vertices
i.e. $\alpha(G) + \beta (G) = n$
2) $\alpha(G) \geq \frac{n}{\chi (G)}$
Since , for planar graphs , $\chi (G) \leqslant 4$ . So, $\frac{n}{\alpha (G)} \leqslant 4$. Therefore, $\frac{n}{4} \leqslant \alpha (G)$.
Now, Since, $\frac{n}{4} \leqslant \alpha (G)$. So, $\;\frac{n}{4} \leqslant (n-\beta(G))$. Therefore, $\beta(G) \leqslant \frac{3n}{4}$.
So, both things should be correct in case of minimum vertex cover and largest independent set not only vertex cover and independent set.