<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] Divide numerator and denominator by $a$ (where $a>0$). The denominator becomes
$$\frac{1}{a}\sqrt{a^2-b^2}=\sqrt{\frac{1}{a^2}}\times\sqrt{a^2-b^2}=\sqrt{1-\frac{b^2}{a^2}}.$$
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