How do I rearrange this? Hi please could some one explain the steps taken to rearange the below. I've had a dabble but don't seem to be getting it.
$$\frac{C}{\sqrt{C^2 - V^2}} $$
To
$$\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}$$
Any help would be greatly appreciated. 
Thanks
 A: Assuming $C > 0$ then
$\frac C{\sqrt{C^2 - V^2}}=$
$\frac {C*\frac 1C}{(\sqrt{C^2 - V^2})\frac 1C}=$
$\frac {1}{(\sqrt{C^2 - V^2})\frac 1{\sqrt{C^2}})}=$
$\frac {1}{\sqrt{(C^2 - V^2)\frac 1{C^2}}}=$
$\frac {1}{\sqrt{\frac {C^2}{C^2} - \frac {V^2}{C^2}}}=$
$\frac 1{\sqrt{1 - \frac {V^2}{C^2}}}$
.... or ...
$\frac {C}{\sqrt {C^2 - V^2}}=$
$\frac {C}{\sqrt{C^2 - C^2\frac {V^2}{C^2}}}=$
$\frac {C}{\sqrt{C^2(1 - \frac {V^2}{C^2})}}=$
$\frac {C}{\sqrt{C^2}\sqrt{1-\frac {V^2}{C^2}}}=$
$\frac {C}{C\sqrt{1-\frac {V^2}{C^2}}}=$
$\frac 1 {\sqrt{1-\frac {V^2}{C^2}}}$
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Addendum.  It should be pointed out that if $C < 0$ then $\sqrt{C^2} \ne C$ but $\sqrt{C^2} = |C| = -C> 0$.  And $C = -\sqrt{C^2}$.  Thus if $C < 0$ then $\frac C{\sqrt{C^2 - V^2}} = \frac {C*\frac 1C}{\sqrt{C^2 -V^2}\frac 1{-\sqrt{C^2}}} = -\frac 1{\sqrt{1 - \frac{V^2}{C^2}}}$ 
This also requires that $C \ne 0$ because we can't divide by $0$.  However for $\frac C{\sqrt{C^2 - V^2}}$ to be defined then $\sqrt{C^2 - V^2} \ne 0$ which means $C^2 \ne V^2$ and $C^2 - V^2 \ge 0$ which means $C^2 \ge V^2$ so $C^2 > V^2$ and $V^2 \ge 0$ so $C^2 > 0$ so $C \ne 0$.
But we must assume $|C| > |V|$.
A: $$
\begin{align}
\frac{C}{\sqrt{C^2 - V^2}}
&=\frac{C}{\sqrt{C^2 - V^2}}\cdot\frac{1/C}{1/C}\\
&=\frac{C/C}{\frac{1}{C}\sqrt{C^2\bigg(1 - \frac{V^2}{C^2}\bigg)}}\\
&=\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\frac{C}{C}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.
\end{align}
$$
This result is only true for $C\gt0$. For $C\lt0$, we would get a slightly different answer:
$$\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{\frac{|C|}{C}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{-\sqrt{1 - \frac{V^2}{C^2}}}=-\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.$$
A: If $C>0,$
$$\frac{C}{\sqrt{C^2-V^2}} = \frac{C}{\sqrt{C^2\left(1-\frac{V^2}{C^2}\right)}} = \frac{C}{\sqrt{C^2}\sqrt{\left(1-\frac{V^2}{C^2}\right)}} $$
$$= \frac{C}{C\sqrt{\left(1-\frac{V^2}{C^2}\right)}}= \frac{1}{\sqrt{\left(1-\frac{V^2}{C^2}\right)}}.$$
