Proving a Complex number equality

To Prove: If $\displaystyle p=\operatorname{cis}\theta =\cos\theta+i\sin\theta$ and $\displaystyle q=\operatorname{cis}\phi =\cos\phi+i\sin\phi$, then show that

$\displaystyle \frac{(p+q)(pq-1)}{(p-q)(pq+1)}=\frac{\sin\theta+\sin\phi}{\sin\theta-\sin\phi}$

My Attempt: $\displaystyle p=e^{i\theta}$ and $\displaystyle q=e^{i\phi}$

Then we have

$\displaystyle \frac{(e^{i\theta}+e^{i\phi})(e^{i(\theta+\phi)}-e^{i0})}{(e^{i\theta}-e^{i\phi})(e^{i(\theta+\phi)}+e^{i0})}$

I think I am going in the wrong direction here. But putting the full $\cos$ and $\sin$ in the problem will just blow the size of the expression. Any hints ?

Answer

Starting from the expression which you have just derived, we can remain in the complex exponential domain and simplify further in this domain as follows:-
$$\displaystyle \frac{(e^{i\theta}+e^{i\phi})(e^{i(\theta+\phi)}-e^{i0})}{(e^{i\theta}-e^{i\phi})(e^{i(\theta+\phi)}+e^{i0})}=\frac{e^{i(2\theta+\phi)}-e^{i\theta}+e^{i(\theta+2\phi)}-e^{i\phi}}{e^{i(2\theta+\phi)}+e^{i\theta}-e^{i(\theta+2\phi)}-e^{i\phi}}\\=\frac{e^{i(\theta+\phi)}(e^{i\theta}-e^{-i\theta}+e^{i\phi}-e^{-i\phi})}{e^{i(\theta+\phi)}(e^{i\theta}-e^{-i\theta}-e^{i\phi}+e^{-i\phi})}\\=\frac{\color{blue}{(e^{i\theta}-e^{-i\theta})}+\color{red}{(e^{i\phi}-e^{-i\phi})}}{\color{blue}{(e^{i\theta}-e^{-i\theta})}-\color{red}{(e^{i\phi}+e^{-i\phi})}}\\=\frac{\color{blue}{2i\sin\theta}+\color{red}{2i\sin\phi}}{\color{blue}{2i\sin\theta}-\color{red}{2i\sin\phi}}\\=\frac{\sin\theta+\sin\phi}{\sin\theta-\sin\phi}$$