How to transform this equation? Let

and

I am reading a paper where the author states that in terms of the coordinates ($\theta$, $\phi$) the second equation is translated to

Does anyone know how this happens?
Thanks.
 A: The approach (I am not writing down all the steps of the calculation in detail, but I have tried everything out and it works; so, if you ask, I can write down every calculation involved in case you get stuck somewhere. But it would be great if you could try out the calculations yourself. I'll try to leave the answer at a stage from where it will be easy for you to pick up.) - 
We are given that $\zeta=\cot{\dfrac{\theta}{2}}e^{i\phi}$ and $\zeta=\dfrac{x\cdot y}{z}$, where $x=e^{i\phi_0}, y=\cot{\dfrac{\theta_0}{2}}+e^{i\gamma}\cot{\dfrac{\Theta}{2}}$ and $z=1-e^{i\gamma}\cot{\dfrac{\theta_0}{2}}\cot{\dfrac{\Theta}{2}}$.
(i) The first part i.e. deriving the expression for $\cos{\theta}$ - 
$\cot{\dfrac{\theta}{2}}e^{i\phi}=\dfrac{e^{i\phi_0}\cdot y}{z}$
$e^{i(\phi-\phi_0)}=\dfrac{y\cdot \tan{\dfrac{\theta}{2}}}{z}$
Also, $\cot{\dfrac{\theta}{2}}e^{-i\phi}=\dfrac{e^{-i\phi_0}\cdot \bar{y}}{\bar{z}}$ (taking conjugate of the first expression as given)
$e^{i(\phi-\phi_0)}=\dfrac{\cot{\dfrac{\theta}{2}}\cdot \bar{z}}{\bar{y}}$
Therefore - 
$\dfrac{\cot{\dfrac{\theta}{2}}\cdot \bar{z}}{\bar{y}}=\dfrac{y\cdot \tan{\dfrac{\theta}{2}}}{z}$
$\cot^2{\dfrac{\theta}{2}}=\dfrac{y\bar{y}}{z\bar{z}}$
$\dfrac{1+\cos{\theta}}{1-\cos{\theta}}=\dfrac{|y|^2}{|z|^2}$. I would suggest you to try it out from this point onwards.
(ii) The second part i.e. deriving the expression for $\tan{\phi}$ - 
Now, $\tan{\phi}=\dfrac{Im(\zeta\tan{\dfrac{\theta}{2}})}{Re(\zeta\tan{\dfrac{\theta}{2}})}=\dfrac{Im(\zeta)}{Re(\zeta)}=\dfrac{\dfrac{\zeta-\bar{\zeta}}{2i}}{\dfrac{\zeta+\bar{\zeta}}{2}}=\dfrac{i(\bar{\zeta}-\zeta)}{\bar{\zeta}+\zeta}=\dfrac{i(\dfrac{\bar{x}\bar{y}}{\bar{z}}-\dfrac{xy}{z})}{\dfrac{\bar{x}\bar{y}}{\bar{z}}+\dfrac{xy}{z}}$
$\tan{\phi}=\dfrac{i(\bar{x}\bar{y}z-xy\bar{z})}{\bar{x}\bar{y}z+xy\bar{z}}$. Now put the expressions for $x,y$ and $z$ and simplify.
