How does this transform algebraically

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I have two lines of working that I am trying to understand.

First line: \begin{equation} \frac{(1+Z^{-1})\tan\frac{wc}{2}}{(1-Z^{-1})+(1+Z^{-1})\tan\frac{wc}{2}} \end{equation}
Next line: \begin{equation} \frac{(1+Z^{-1})\tan\frac{wc}{2}}{(1+\tan\frac{wc}{2})-(1-\tan\frac{wc}{2})Z^{-1}} \end{equation}

I know that they equal each other but I'm not sure how the denominator is transformed from one to the other. Am I missing some algebra trick?

Thanks for your help.

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You can do the following.

$(1-z^{-1})+(1+z^{-1})\tan \frac{wc}{2} =1-z^{-1}+\tan \frac{wc}{2}+z^{-1}\tan \frac{wc}{2}=\\(1+\tan \frac{wc}{2})-z^{-1}+z^{-1}\tan \frac{wc}{2}=(1+\tan \frac{wc}{2})+(\tan \frac{wc}{2}-1)z^{-1}=\\(1+\tan \frac{wc}{2})-(1-\tan \frac{wc}{2})z^{-1} $

I've tried to do it in detail. There isnøt really any trick to it.