Alan Smith

2022-01-03

How to prove $\frac{{e}^{jx}-j{e}^{-jx}}{j{e}^{jx}-{e}^{-jx}}=\frac{\mathrm{tan}x-1}{\mathrm{tan}x+1}$
where $j=\sqrt{-1}$
Ive

jgardner33v4

It can be done easily from the left hand side itself, if you use Eulers

nghodlokl

Assuming $j=\sqrt{-1}$
First of all, observe that:
$\frac{\mathrm{tan}x-1}{\mathrm{tan}x+1}=\frac{\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{cos}x}}{\frac{\mathrm{sin}x+\mathrm{cos}x}{\mathrm{cos}x}}=\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}$
Moving to complex numbers:
$\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}=\frac{{e}^{jx}-{e}^{-jx}-j\left({e}^{jx}+{e}^{-jx}\right)}{{e}^{jx}-{e}^{-jx}+j\left({e}^{jx}+{e}^{-jx}\right)}=\frac{\left(1-j\right){e}^{jx}-\left(1+j\right){e}^{-jx}}{\left(1+j\right){e}^{jx}-\left(1-j\right){e}^{-jx}}=$
$\frac{\left(1-j\right)\left({e}^{jx}-\frac{1+j}{1-j}{e}^{-jx}\right)}{\left(1-j\right)\left(\frac{1+j}{1-j}{e}^{jx}-{e}^{-jx}\right)}=\frac{{e}^{jx}-\frac{1+j}{1-j}{e}^{-jx}}{\frac{1+j}{1-j}{e}^{jx}-{e}^{-jx}}$
Now, observe that:
$\frac{1+j}{1-j}=\frac{{\left(1+j\right)}^{2}}{\left(1-j\right)\left(1+j\right)}=\frac{1+{j}^{2}+2j}{{1}^{2}-{j}^{2}}=\frac{1-1+2j}{1-\left(-1\right)}=j$
Hence:
$\frac{\mathrm{tan}x-1}{\mathrm{tan}x+1}=\frac{{e}^{jx}-j{e}^{-jx}}{j{e}^{jx}-{e}^{-jx}}$

Vasquez

Note that:
$RHS=\frac{\mathrm{tan}x-\mathrm{tan}\frac{\pi }{4}}{1+\mathrm{tan}x\mathrm{tan}\frac{\pi }{4}}=\mathrm{tan}\left(x-\frac{\pi }{4}\right)=\frac{\mathrm{sin}\left(x-\frac{\pi }{4}\right)}{\mathrm{cos}\left(x-\frac{\pi }{4}\right)}=\frac{\frac{1}{2j}}{\frac{1}{2}}\frac{{e}^{j\left(x-\frac{\pi }{4}\right)}-{e}^{-j\left(x-\frac{\pi }{4}\right)}}{{e}^{j\left(x-\frac{\pi }{4}\right)}+{e}^{-j\left(x-\frac{\pi }{4}\right)}}$
and since ${e}^{\frac{j\pi }{4}}=\mathrm{cos}\left(\frac{\pi }{4}\right)+j\mathrm{sin}\left(\frac{\pi }{4}\right),{e}^{\frac{j\pi }{4}}=\mathrm{cos}\left(\frac{\pi }{4}\right)-j\mathrm{sin}\left(\frac{\pi }{4}\right)$ by Euler's formula:
$\frac{\frac{1}{2j}}{\frac{1}{2}}\frac{\left(\mathrm{cos}\frac{\pi }{4}-j\mathrm{sin}\frac{\pi }{4}\right){e}^{jx}-\left(\mathrm{cos}\frac{\pi }{4}+j\mathrm{sin}\frac{\pi }{4}\right){e}^{-jx}}{\left(\mathrm{cos}\frac{\pi }{4}-j\mathrm{sin}\frac{\pi }{4}\right){e}^{jx}+\left(\mathrm{cos}\frac{\pi }{4}+j\mathrm{sin}\frac{\pi }{4}\right){e}^{-jx}}$

$=\frac{\left(1-j\right){e}^{jx}-\left(1+j\right){e}^{-jx}}{\left(j+1\right){e}^{jx}+\left(j-1\right){e}^{-jx}}$
and dividng top and bottom by $1-j,\frac{1+j}{1-j}\frac{1+j}{1+j}=\frac{2j}{j}=j$ and $\frac{j-1}{1-j}=\frac{-\left(1-j\right)}{1-j}=-1$,so we obtain $\frac{{e}^{jx}-j{e}^{-jx}}{j{e}^{jx}-{e}^{-jx}}$ which is the LHS.

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