# Simplifying an expression: Stuck with it

I have to prove that the expression

$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$

is equal to

$$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$

where $$\omega_r= \frac{1}{\sqrt{LC}}$$.

My attempt:

What I started to do was to get rid of the denominators in the fraction and put everything together.

$$\frac{\omega^2C^2L-C}{\omega^2C^2L-C+\omega^2CL^2-L}$$

Then I divided the denominator by the numerator

$$\frac{1}{1+\frac{\omega^2CL^2-L}{\omega^2C^2L-C}}$$

And I'm kind of stuck now. Can someone give an hint on how should I proceed next? Or is there any easier way to start the proof? I'm just looking for a hint, thanks.

• Where do these expressions come from? Why do you believe they should be equal? Desmos suggests they aren't equal, in general. – Robert Howard Nov 3 '18 at 5:26

The expressions are NOT equal when $$\omega = 2$$ and $$L=1$$ and $$C=1$$:
$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}} \;\; = \;\; \frac{2\cdot1\; - \; \frac{1}{2\cdot 1}}{2\cdot 1 \; - \; \frac{1}{2\cdot1} \; + \; 2\cdot1 \; - \; \frac{1}{2\cdot 1}} \;\; = \;\; \frac{\frac{3}{2}}{\;4 - 1\;} \;\; = \;\; \frac{1}{2}$$
$$\frac{1}{3 \; - \; \left[ \left(\frac{\omega_r}{\omega}\right)^2 + \left(\frac{\omega}{\omega_r}\right)^2\right]} \; = \; \frac{1}{3 \; - \; \left[ \left(\frac{1}{2}\right)^2 + \left(\frac{2}{1}\right)^2\right]} \; = \; \frac{1}{\;3 \; - \; \frac{1}{4} \; - \; 4\;} \; = \; \frac{1}{\;-\frac{5}{4}\;} \; = \; -\frac{4}{5},$$
where I've used the fact that $$\;\omega_r= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1\cdot 1\;}} = 1.$$