Proof of trigonometric identity $\frac{\cos x+i\sin x+1}{\cos x+i\sin x-1}= -\frac{i}{\tan \frac{x}{2}}$ I was given a task of proving the following identity:
$$\frac{\cos x+i\sin x+1}{\cos x+i\sin x-1}= -\frac{i}{\tan \frac{x}{2}}$$
I am not looking for a solution, just some kind of a hint to start off. Thanks.
 A: HINT:
Use the identity
$$\cos x +i\sin x=e^{ix}$$
and multiply numerator and denominator by $e^{-ix/2}$. Furthermore, you need
$$\cos x=\frac12 (e^{ix}+e^{-ix})\\
\sin x=\frac{1}{2i} (e^{ix}-e^{-ix})$$
A: Hint: write everything in exponential form.
Now, the following identities are useful:


*

*$e^{ix} \equiv \cos(x)=i\sin(x)$

*$\tan(\frac{x}{2}) \equiv \frac{i\left[e^{-i\left(\frac{x}{2}\right)}-e^{i\left(\frac{x}{2} \right)}\right]}{e^{i\left( \frac{x}{2}\right)}+e^{-i\left(\frac{x}{2} \right)}}$

A: $$e^{ix} = cos(x) + isin(x)$$
Where $i$ is the imaginary constant.
Now go derive formulas for sin(x), cos(x) and furthermore use those formulas to derive an expression for tan(x) and set that equal to what you have above and see if you can get 0 = 0 or something like that after simplifying
A: Observe that $$(\cos y+i\sin y)(\cos y+i\sin y)=\cdots=\cos2y+i\sin2y$$
and $$(\cos y+i\sin y)(\cos y-i\sin y)=\cdots=1$$
$$\implies\frac{\cos y+i\sin y}{\cos y-i\sin y}=\cos2y+i\sin2y$$
$$\implies\frac{\cos2y+i\sin2y}1=\frac{\cos y+i\sin y}{\cos y-i\sin y}$$
Dividing the numerator & the denominator of the Right Hand Side by $i\cos y$
$$\frac{\cos2y+i\sin2y}1=\frac{-i+\tan y}{-i-\tan y}$$
Apply Componendo and dividendo
