trig representation

[Arno K]

I made a plot of both, they seem to be the same, now I start trying by Hand.
Arno
P.S. If you want it quick and dirty: take the result from the Prime and yours and make an equation, use Binoms in the calc-result and get out the 2 from the exponents, multiply by 4 and take the resultimg two in the ln of your result, omit ln from both sides, you should now have \[abs(\frac{sin(z)+1}{sin(z)-1})=abs(\frac{(1+sin(z))^2}{(cos(z))^2})\] Put the abs on both sides on denominator as well as nominator, multiply crosswise and divide by \(abs(sin(z)+1)\), you have now \[abs((cos(z)^2)=abs((1+sin(z))*(sin(z)-1))\]Inside abs sin(z)-1 is the same as 1-sin(z) and because of \(sin(z)^2+cos(z)^2=1\) you are done. But how to make the calc show that, I don't know as I usually do things like that by hand.
Arno