The question I met is to show that if $z=\cos (\theta)+i\sin(\theta)$ with $i=\sqrt{-1}$,then $ Re(\frac{z-1}{z+1})=0$
In the normal way, we found that: $$\frac{z-1}{z+1}=\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}\\ =\frac{\bigl(\cos(\theta)-1+i\sin(\theta)\bigl)\bigl(\cos(\theta)+1-i\sin(\theta)\bigl)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\ =\frac{\cos^2(\theta)+\cos(\theta)-\cos(\theta)-1+\sin^2(\theta)+2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\ =\frac{2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}$$
So $Re(\frac{z-1}{z+1})=0$
If we do it in another way: $$ \frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}=\frac{-2\sin^2(\frac{\theta}{2})+2i\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})}{2cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\frac{\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\frac{\cos\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)+i\sin\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\bigl(\cos(-\theta-\frac{\pi}{2})+i\sin(-\theta-\frac{\pi}{2})\bigl) $$
So the real part of it will be $-\tan(\frac{\theta}{2})\cos(-\theta-\frac{\pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
You should be more careful when applying transformations of sines into cosines.
It's much easier than that: $$ \sin\frac{\theta}{2}-i\cos\frac{\theta}{2}= i\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right) $$ You could write this in trigonometric form $$ =\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) \left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right) =\cos\left(\frac{\theta}{2}+\frac{\pi}{2}\right)+ i\sin\left(\frac{\theta}{2}+\frac{\pi}{2}\right) $$ and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=\frac{z-1}{z+1}$ is $$ \bar{w}=\frac{\bar{z}-1}{\bar{z}+1} $$ but, since $|z|=1$, we have $\bar{z}=z^{-1}$; therefore $$ \bar{w}=\frac{z^{-1}-1}{z^{-1}+1}=\frac{1-z}{1+z}=-w $$ From $\bar{w}=-w$ it follows that $\operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $\theta$, you can consider $z=u^2$, where $u=\cos(\theta/2)+i\sin(\theta/2)$; then $$ w=\frac{z-1}{z+1}=\frac{u^2-1}{u^2+1}=\frac{u-u^{-1}}{u+u^{-1}} =\frac{u-\bar{u}}{u+\bar{u}}= \frac{2i\sin(\theta/2)}{2\cos(\theta/2)}=i\tan\frac{\theta}{2} $$