<s>[INST] 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
 [/INST] $$
\begin{align}
\frac{C}{\sqrt{C^2 - V^2}}
&=\frac{C}{\sqrt{C^2 - V^2}}\cdot\frac{1/C}{1/C}\\
&=\frac{C/C}{\frac{1}{C}\sqrt{C^2\bigg(1 - \frac{V^2}{C^2}\bigg)}}\\
&=\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\frac{C}{C}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.
\end{align}
$$
This result is only true for $C\gt0$. For $C\lt0$, we would get a slightly different answer:
$$\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{\frac{|C|}{C}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{-\sqrt{1 - \frac{V^2}{C^2}}}=-\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.$$
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