If $\frac{1}{b} \div \frac{b}{a} = \frac{a^2}{b}$, where a, b not equal to 0, then find the value of $\frac{\frac{a}{(\frac{1}{b})} - 1}{\frac{a}{b}}$.

Given :

$ \frac{1}{b} \div \frac{b}{a} =\frac{a^{2}}{b}$ , a, b not equal to 0.

To find :

We have to find the value of $\frac{\frac{a}{(\frac{1}{b})} - 1}{\frac{a}{b}}$.

Solution :

$\frac{1}{b} \times \frac{a}{b} =\frac{a^{2}}{b}$

$a\times b=a^{2} \times b^{2}$

$ab=a^{2} b^{2}$

$ab( ab-1) =0$

$ab=0\ or\ ab=1$

$ab \ not \ equal \ to \ 0 $ [since a not equal to 0 and b not equal to 0]

Therefore,

$ab = 1.$

$\frac{\frac{a}{1/b} -1}{\frac{a}{b}} \ =\ \frac{\frac{a-\left(\frac{1}{b}\right)}{1/b}}{a/b}$

 $=\frac{a-\left(\frac{1}{b}\right)}{1/b} \times \frac{b}{a}$

$ =\ \frac{( ab-1) /b}{1/b} \times \frac{b}{a}$

$ =\ \frac{b( ab-1)}{a}$

$ =\frac{b( 1-1)}{a}$

$=0$

Therefore, the value of $\frac{\frac{a}{(\frac{1}{b})} - 1}{\frac{a}{b}}$ is $0$.