<s>[INST] Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left part, and I ask for the explaining of this. Thanks a lot!
 [/INST] We can prove this in a reverse way
Lets start with $$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
$$=\frac{1/a}{\sqrt{\frac{a^2-b^2}{a^2}}}$$
$$=\frac{1/a}{\frac{\sqrt{a^2-b^2}}{a}}$$
$$=\frac{\frac{1}{a}(a)}{\sqrt{a^2-b^2}}$$
$$=\frac{1}{\sqrt {a^2 - b^2}}$$
Hence,$$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}=\frac{1}{\sqrt {a^2 - b^2}}$$
which is nothing but,
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
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