Prove that \(\sqrt{1+\mathrm{cosec}(\theta/2)}=\left(1-e^{i\theta}\right)^{-1/2}+\left(1-e^{-i\theta}\right)^{-1/2}\).

Step 1: \(1-e^{i\theta}=e^{i\theta/2}\left(e^{-i\theta/2}-e^{i\theta/2}\right)=e^{i\theta/2}\left(-2i\sin\frac\theta2\right)=2\sin\frac\theta2\,e^{i\left(\frac\theta2-\frac\pi2\right)}\) (since \(-i=e^{-i\pi/2}\)).

\(\therefore\left(1-e^{i\theta}\right)^{-1/2}=\left(2\sin\frac\theta2\right)^{-1/2}e^{i\frac{\pi-\theta}{4}}\), and similarly (conjugate) \(\left(1-e^{-i\theta}\right)^{-1/2}=\left(2\sin\frac\theta2\right)^{-1/2}e^{-i\frac{\pi-\theta}{4}}\) (principal values, \(0<\theta<2\pi\)).

Step 2: RHS \(=\left(2\sin\frac\theta2\right)^{-1/2}\left(e^{i\frac{\pi-\theta}{4}}+e^{-i\frac{\pi-\theta}{4}}\right)=\frac{2\cos\frac{\pi-\theta}{4}}{\sqrt{2\sin\frac\theta2}}\).

Step 3: Squaring: \(\text{RHS}^2=\frac{4\cos^2\frac{\pi-\theta}{4}}{2\sin\frac\theta2}=\frac{2\left[1+\cos\frac{\pi-\theta}{2}\right]}{2\sin\frac\theta2}=\frac{1+\sin\frac\theta2}{\sin\frac\theta2}=1+\mathrm{cosec}\frac\theta2\).

Since RHS > 0, \(\left(1-e^{i\theta}\right)^{-1/2}+\left(1-e^{-i\theta}\right)^{-1/2}=\sqrt{1+\mathrm{cosec}\frac\theta2}\) (proved).