I have the function $f(n)= \pi (n+\frac{1}{2})+\epsilon \frac{2\pi(n+1/2)-a}{2\pi(n+1/2)}$ when $\epsilon<<1$, I can't understand what identity I need to use to prove that:

$\tan(f(n))\approx \frac{2\pi(n+1/2)}{\epsilon(2\pi(n+1/2)-a)}$ I tried with taylor but the first element in $f(x)$ taking tan function to $\infty$

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    $\begingroup$ Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$? $\endgroup$ – Calvin Godfrey Jan 11 at 13:28
  • $\begingroup$ What if $a\approx2\pi(n+1/2)$? $\endgroup$ – Barry Cipra Jan 11 at 13:37

The first step is to use

$$ \tan(x +n\pi +\pi/2) = -\frac{\cos x}{\sin x}$$

The following is easy


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