Ratio of two tanned alternate signed even-odd binomial summation

If you visit the wiki: Expansions of certain Trigonometric Functions , you'll find this and it's proof:

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$\large{\tan(n\theta) = \frac{\displaystyle \sum_{\substack{r=0 \\ 2r+1 \leq n}} (-1)^r \binom{n}{2r+1} \tan^{2r+1}(\theta)}{\displaystyle \sum_{\substack{r=0 \\ 2r \leq n}} (-1)^r \binom{n}{2r} \tan^{2r}(\theta)}}$

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Substituting the required values, we can obtain our answer as $-0.577$ .

Well, it is just as easy as you would think.

The upper sum can be written as -

$\displaystyle \frac{1}{\iota} \sum_{r=1}^{2k}{\binom{4k}{2r-1} {(\iota tan \theta)}^{2r-1}}$

And the lower sum as -

$\displaystyle \sum_{r=0}^{2k}{\binom{4k}{2r} {(\iota tan \theta)}^{2r}}$

Saw something?

They both add to(if we neglect that $\frac{1}{\iota}$ term) - ${(1+\iota tan\theta)}^{4k}$

So,

$\displaystyle R = \frac{\frac{1}{\iota} \iota Im({(1+\iota tan\theta)}^{4k})}{Re({(1+\iota tan\theta)}^{4k})}$

Using Euler's formula, we get

$\displaystyle R\left(k, \theta\right) = tan\left(4k\theta\right)$

Substituting values we get

$\displaystyle R\left(100, \frac{\pi}{480}\right) = -\frac{1}{\sqrt{3}}$