Suppose $\gcd(\deg(f),\deg (g))=1$. Show that $g(x)$ is irreducible in $k(\alpha)[X]$.
Here is #1: Let $h$ denote the minimal polynomial of $\beta$ in $k(\alpha)[x]$, then $$ [k(\alpha,\beta):k(\alpha)] = \deg(h) $$ But $[k(\alpha):k] = \deg(f)$, so $$ [k(\alpha,\beta):k] = \deg(f)\deg(h) $$ But $[k(\alpha,\beta):k] = [k(\alpha):k][k(\beta):k] = \deg(f)\deg(g)$ since the two field extensions have degrees that are relatively prime. Hence, $$ \deg(h) = \deg(g) $$ But $h\mid g$ in $k(\alpha)[x]$, so $h=g$ must hold.