The following answer was communicated to me by Keith Conrad:

See:

>M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25
>(1970), 638--641.

>http://math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are
relatively prime, then
$K(a,b)$ = $K(a+b)$, which answers the students question in the
affirmative.  His proof shows the same conclusion holds under the
weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees
only to get that degree formula above, which can occur even in cases
where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.