# How to prove \frac{e^{jx}-je^{-jx}}{je^{jx}-e^{-jx}}=\fra

How to prove $$\displaystyle{\frac{{{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{j}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}={\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}$$
where $$\displaystyle{j}=\sqrt{{-{1}}}$$
I've tried to prove it from the right hand
$$\displaystyle{\tan{{x}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}$$ (2)
and thus
$$\displaystyle{\tan{{x}}}-{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}$$ (3)
and
$$\displaystyle{\tan{{x}}}+{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}$$ (4)
The right side of Eq. (1) can be written as
$$\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{e}^{{{j}{x}}}+{j}{e}^{{-{j}{x}}}}}}$$ (5)
Compared with the left side of Eq. (1), the numerator has the same part $$\displaystyle{e}^{{{j}{x}}}−{j}{e}^{{−{j}{x}}}$$, and the denominator has the same part $$\displaystyle{j}{e}^{{{j}{x}}}−{e}^{{-{j}{x}}}$$, but the other parts don't seem to be zero constantly.
I've checked Eqs. (1) and (5), they are all correct numerically. Can anyone help to find where I am wrong? Thanks in advance!

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jgardner33v4
It can be done easily from the left hand side itself, if you use Euler's formula:
$$\displaystyle{e}^{{{j}{x}}}={\cos{{x}}}+{j}{\sin{{x}}}$$
Just substitute this in LHS and simplify.
###### Not exactly what you’re looking for?
nghodlokl
Assuming $$\displaystyle{j}=\sqrt{{-{1}}}$$
First of all, observe that:
$$\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\cos{{x}}}}}}}}{{{\frac{{{\sin{{x}}}+{\cos{{x}}}}}{{{\cos{{x}}}}}}}}}={\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\sin{{x}}}+{\cos{{x}}}}}}$$
Moving to complex numbers:
$$\displaystyle{\frac{{{\sin{{x}}}-{\cos{{x}}}}}{{{\sin{{x}}}+{\cos{{x}}}}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}={\frac{{{\left({1}-{j}\right)}{e}^{{{j}{x}}}-{\left({1}+{j}\right)}{e}^{{-{j}{x}}}}}{{{\left({1}+{j}\right)}{e}^{{{j}{x}}}-{\left({1}-{j}\right)}{e}^{{-{j}{x}}}}}}=$$
$$\displaystyle{\frac{{{\left({1}-{j}\right)}{\left({e}^{{{j}{x}}}-{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{-{j}{x}}}\right)}}}{{{\left({1}-{j}\right)}{\left({\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}\right)}}}}={\frac{{{e}^{{{j}{x}}}-{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{-{j}{x}}}}}{{{\frac{{{1}+{j}}}{{{1}-{j}}}}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}$$
Now, observe that:
$$\displaystyle{\frac{{{1}+{j}}}{{{1}-{j}}}}={\frac{{{\left({1}+{j}\right)}^{{2}}}}{{{\left({1}-{j}\right)}{\left({1}+{j}\right)}}}}={\frac{{{1}+{j}^{{2}}+{2}{j}}}{{{1}^{{2}}-{j}^{{2}}}}}={\frac{{{1}-{1}+{2}{j}}}{{{1}-{\left(-{1}\right)}}}}={j}$$
Hence:
$$\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{j}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}$$
Vasquez

Note that:
$$RHS=\frac{\tan x-\tan \frac{\pi}{4}}{1+\tan x \tan \frac{\pi}{4}}=\tan(x-\frac{\pi}{4})=\frac{\sin(x-\frac{\pi}{4})}{\cos(x-\frac{\pi}{4})}=\frac{\frac{1}{2j}}{\frac{1}{2}}\frac{e^{j(x-\frac{\pi}{4})}-e^{-j(x-\frac{\pi}{4})}}{e^{j(x-\frac{\pi}{4})}+e^{-j(x-\frac{\pi}{4})}}$$
and since $$e^{\frac{j\pi}{4}}=\cos(\frac{\pi}{4})+j\sin(\frac{\pi}{4}),e^{\frac{j\pi}{4}}=\cos(\frac{\pi}{4})-j\sin(\frac{\pi}{4})$$ by Euler's formula:
$$\frac{\frac{1}{2j}}{\frac{1}{2}} \frac{(\cos \frac{\pi}{4}-j \sin \frac{\pi}{4})e^{jx}-(\cos \frac{\pi}{4}+j \sin \frac{\pi}{4})e^{-jx}}{(\cos \frac{\pi}{4}-j \sin \frac{\pi}{4})e^{jx}+(\cos \frac{\pi}{4}+j \sin \frac{\pi}{4})e^{-jx}}$$
$$=\frac{1}{j} \frac{(1-j)e^{jx}-(1+j)e^{-jx}}{(1-j)e^{jx}+(1+j)e^{-jx}} \ \ \ \ \ \ \ \ \ \ (\cos \frac{\pi}{4}=\sin \frac{\pi}{4})$$
$$=\frac{(1-j)e^{jx}-(1+j)e^{-jx}}{(j+1)e^{jx}+(j-1)e^{-jx}}$$
and dividng top and bottom by $$1-j, \frac{1+j}{1-j}\frac{1+j}{1+j}=\frac{2j}{j}=j$$ and $$\frac{j-1}{1-j}=\frac{-(1-j)}{1-j}=-1$$,so we obtain $$\frac{e^{jx}-je^{-jx}}{je^{jx}-e^{-jx}}$$ which is the LHS.