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I have a doubt in complex numbers which I am unable to solve. The question is

Prove that $$\left(\frac{e^{2x}-1}{e^{2x}+1}\right)i=\tan{ix}$$

I tried using hyperbolic sin and cosines but failed. Can anybody guide me how to tackle this question

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$$\tan ix=\frac{\sin ix}{\cos ix}=\frac{\frac{e^{-x}-e^x}{2i}}{\frac{e^{-x}+e^x}{2}}=\frac1i\frac{e^{-x}-e^x}{e^{-x}+e^x}\cdot\frac{e^x}{e^x}=\frac1i\frac{1-e^{2x}}{1+e^{2x}}=$$

$$=i\frac{e^{2x}-1}{e^{2x}+1}$$

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HINT:

Euler Formula says, $e^{iy}=\cos y+i\sin y$

So,$e^{2x}=e^{i(-2ix)}=\cdots$

Then use Tangent half-angle formula

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Hint: You should definitely memorize the following formulas. You doubtless already know their counterparts in terms of $\pm 1$, so they're easy to remember.

$$\sin \pm ix = \pm i \sinh x$$ $$\cos \pm ix = i \cosh x$$ and so $$\tan \pm ix = \pm i \tanh x$$

If you recognize the formulas, these should get you what you need.

In the end, "$i$" pulls out of the circular trig functions, converting them to hyperbolic trig functions.

And, it's obviously reversible.

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