Half Angle Trig and Complex Numbers

Question

By making use of the half-angle formulae, or otherwise, prove that $$\frac{1+\cos x+i\sin x}{1-\cos x+i\sin x}=\cot{\frac x2} e^{i(x-\frac\pi2)}$$

Answer 1

$$\frac{1+\cos2y+i\sin2y}{1-\cos2y+i\sin2y}=\frac{2\cos^2y+2i\sin y\cos y}{2\sin^2y+2i\sin y\cos y}$$

$$=\frac{\cos y(\cos y+i\sin y)}{\sin y(\sin y+i\cos y)}$$

$$=\cot y\frac{\cos y+i\sin y}{\cos\left(\frac\pi2-y\right)+i\sin \left(\frac\pi2-y\right)}$$

$$=\cot y\frac{e^{iy}}{e^{i\left(\frac\pi2-y\right)}}$$

using Euler's Identity $:\cos z+i\sin z=e^{iz}$