How to prove ejx−je−jxjejx−e−jx=tan⁡x−1tan⁡x+1 where j=−1 Ive

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How to prove ejx−je−jxjejx−e−jx=tan⁡x−1tan⁡x+1  where j=−1  Ive

jgardner33v4 · Accepted Answer

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

nghodlokl · Accepted Answer

Assuming j=−1  First of all, observe that:  tan⁡x−1tan⁡x+1=sin⁡x−cos⁡xcos⁡xsin⁡x+cos⁡xcos⁡x=sin⁡x−cos⁡xsin⁡x+cos⁡x  Moving to complex numbers:  sin⁡x−cos⁡xsin⁡x+cos⁡x=ejx−e−jx−j(ejx+e−jx)ejx−e−jx+j(ejx+e−jx)=(1−j)ejx−(1+j)e−jx(1+j)ejx−(1−j)e−jx=    (1−j)(ejx−1+j1−je−jx)(1−j)(1+j1−jejx−e−jx)=ejx−1+j1−je−jx1+j1−jejx−e−jx    Now, observe that:    1+j1−j=(1+j)2(1−j)(1+j)=1+j2+2j12−j2=1−1+2j1−(−1)=j    Hence:    tan⁡x−1tan⁡x+1=ejx−je−jxjejx−e−jx

Vasquez · Accepted Answer

Note that: RHS=tan⁡x−tan⁡π41+tan⁡xtan⁡π4=tan⁡(x−π4)=sin⁡(x−π4)cos⁡(x−π4)=12j12ej(x−π4)−e−j(x−π4)ej(x−π4)+e−j(x−π4) and since ejπ4=cos⁡(π4)+jsin⁡(π4),ejπ4=cos⁡(π4)−jsin⁡(π4) by Euler&#039;s formula: 12j12(cos⁡π4−jsin⁡π4)ejx−(cos⁡π4+jsin⁡π4)e−jx(cos⁡π4−jsin⁡π4)ejx+(cos⁡π4+jsin⁡π4)e−jx =1j(1−j)ejx−(1+j)e−jx(1−j)ejx+(1+j)e−jx          (cos⁡π4=sin⁡π4) =(1−j)ejx−(1+j)e−jx(j+1)ejx+(j−1)e−jx and dividng top and bottom by 1−j,1+j1−j1+j1+j=2jj=j and j−11−j=−(1−j)1−j=−1,so we obtain ejx−je−jxjejx−e−jx which is the LHS.

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

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