How to prove $\frac{e^{jx} - je^{-jx}}{je^{jx}-e^{-jx}} = \frac{\tan x-1}{\tan x+1} $ How do I prove this equation:
$$
\frac{e^{jx} - je^{-jx}}{je^{jx}-e^{-jx}} = \frac{\tan x-1}{\tan x+1} \tag{1}
$$
where $j=\sqrt{-1}$.
I've tried to prove it from the right hand
$$
\tan x = \frac{e^{jx}-e^{-jx}}{j(e^{jx}+e^{-jx})}\tag{2}
$$
and thus
$$
\tan x - 1 = \frac{e^{jx}-e^{-jx} - j(e^{jx}+e^{-jx})}{j(e^{jx}+e^{-jx})}\tag{3}
$$
and
$$
\tan x + 1 = \frac{e^{jx}-e^{-jx} + j(e^{jx}+e^{-jx})}{j(e^{jx}+e^{-jx})}\tag{4}
$$
The right side of Eq. (1) can be written as
$$
\frac{\tan x-1}{\tan x+1}=\frac{e^{jx}-e^{-jx} - je^{jx}- je^{-jx}} {e^{jx}-e^{-jx} + je^{jx}+je^{-jx}} \tag{5}
$$
Compared with the left side of Eq. (1), the numerator has the same part $e^{jx} - je^{-jx}$, and the denominator has the same part $je^{jx}-e^{-jx}$, 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!
 A: It can be done easily from the left hand side itself, if you use Euler's formula:
$$e^{jx}=\cos x+j\sin x$$
Just substitute this in LHS and simplify.
A: Assuming $j= \sqrt{-1}$...
First of all, observe that:
$$\frac{\tan x-1}{\tan x+1} = \frac{\frac{\sin x - \cos x}{\cos x}}{\frac{\sin x + \cos x}{\cos x}} = \frac{\sin x - \cos x}{\sin x + \cos x}.$$
Moving to complex numbers:
$$\frac{\sin x - \cos x}{\sin x + \cos x} = \frac{e^{jx}-e^{-jx}-j(e^{jx}+e^{-jx})}{e^{jx}-e^{-jx}+j(e^{jx}+e^{-jx})} = \frac{(1-j)e^{jx}-(1+j)e^{-jx}}{(1+j)e^{jx}-(1-j)e^{-jx}} = \\
\frac{(1-j)\left(e^{jx}-\frac{1+j}{1-j}e^{-jx}\right)}{(1-j)\left(\frac{1+j}{1-j}e^{jx}-e^{-jx}\right)} =  \frac{e^{jx}-\frac{1+j}{1-j}e^{-jx}}{\frac{1+j}{1-j}e^{jx}-e^{-jx}}.$$
Now, observe that:
$$\frac{1+j}{1-j} = \frac{(1+j)^2}{(1-j)(1+j)} = \frac{1+j^2+2j}{1^2 - j^2} = \frac{1-1+2j}{1-(-1)} = j.$$
Hence:
$$\frac{\tan x-1}{\tan x+1}  = \frac{e^{jx} - je^{-jx}}{je^{jx}-e^{-jx}}.$$
A: Note that:
$$\text{RHS} = \frac{\tan x - \tan \pi/4}{1 + \tan x \tan \pi/4}= \tan(x - \pi/4) = \frac{\sin(x - \pi/4)}{\cos(x - \pi/4)} = \frac{1/(2j)}{1/2} \frac{e^{j(x-\pi/4)}-e^{-j(x-\pi/4)}}{e^{j(x-\pi/4)}+e^{-j(x-\pi/4)}}.$$
and since $e^{j \pi/4} = \cos(\pi/4) + j \sin(\pi/4), e^{-j \pi/4} = \cos(\pi/4) - j \sin(\pi/4)$ by Euler's formula:
$$\frac{1/(2j)}{1/2} \frac{(\cos \pi/4 - j \sin \pi/4)e^{jx}-(\cos \pi/4 + j \sin \pi/4)e^{-jx}}{(\cos \pi/4 - j \sin \pi/4)e^{jx}+(\cos \pi/4 + j \sin \pi/4)e^{-jx}}$$
$$=\frac{1}{j} \frac{(1 - j)e^{jx}-(1 + j)e^{-jx}}{(1-j)e^{jx}+(1+j)e^{-jx}} \tag{$\cos \pi/4 = \sin \pi/4$}$$
$$= \frac{(1 - j)e^{jx}-(1 + j)e^{-jx}}{(j+1)e^{jx}+(j-1)e^{-jx}}$$
and dividng top and bottom by $1-j$, $\frac{1+j}{1-j} \frac{1+j}{1+j} = \frac{2j}{2}=j$ 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.
