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!
 A: Divide numerator and denominator by $a$ (where $a>0$). The denominator becomes
$$\frac{1}{a}\sqrt{a^2-b^2}=\sqrt{\frac{1}{a^2}}\times\sqrt{a^2-b^2}=\sqrt{1-\frac{b^2}{a^2}}.$$
A: For nonnegative $u$ and $v$, you can write $\sqrt{uv}=\sqrt{u}\cdot\sqrt{v}$. You also have, for any $u$, $\sqrt{u^2}=|u|$. Use this here:
Assume $a^2\ge b^2$ and $a\ne0$, then
$$\sqrt{a^2-b^2}=\sqrt{a^2\left(1-\frac{b^2}{a^2}\right)}=\sqrt{a^2}\cdot\sqrt{1-\frac{b^2}{a^2}}=|a|\sqrt{1-\frac{b^2}{a^2}}$$
If you take the inverse of the preceding, assuming further that $a\ne b$, then
$$\dfrac{1}{\sqrt{a^2-b^2}}=\dfrac{1}{|a|\sqrt{1-\frac{b^2}{a^2}}}$$
Assume further that $a>0$, and
$$\dfrac{1}{\sqrt{a^2-b^2}}=\dfrac{1}{a\sqrt{1-\frac{b^2}{a^2}}}$$
When you take a factor $a^2$ out of a square root, never forget the absolute value unless you are absolutely certain $a\ge0$.
A: I'm assuming you mean "how do we get from the left side of the equation to the right side of the equation. In that case, it's simple division by $a$ in both the numerator and denominator.
This is pretty easy to see in the numerator, but in the denominator, the $a$ is 'absorbed' into the square root as $a^2$. 
A: 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,
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
