Given \(ab = 1\)
Then, \(b = \cfrac{1}{a} = \cfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(\therefore a + b = \cfrac{\sqrt{5}+1}{\sqrt{5}-1} + \cfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(= \cfrac{(\sqrt{5}+1)^2 + (\sqrt{5}-1)^2}{(\sqrt{5})^2 - (1)^2}\)
\(= \cfrac{2[(\sqrt{5})^2 + (1)^2]}{5 - 1}\)
\(= \cfrac{2[5 + 1]}{4}\)
\(= \cfrac{12}{4} = 3\)
Now, \(\left(\cfrac{a}{b} + \cfrac{b}{a}\right)\)
\(= \cfrac{a^2 + b^2}{ab}\)
\(= \cfrac{(a + b)^2 - 2ab}{ab}\)
\(= \cfrac{(3)^2 - 2 \times 1}{1}\)
\(= \cfrac{9 - 2}{1}\)
\(= 7\) (Answer)