Assume \((a+b, ab) = d\), then there is a prime \(p\) that divides \(d\). If \(p|ab\), then according to this, \(p|a\) or \(p|b\).
Assume \(p|a\). If \(p|a\) and \(p|(a+b)\), then \(p|b\).
If \(p\) divides both \(a\) and \(b\), then we have a contradiction (since \((a, b)=1\)).