<s>[INST] How to prove the identity $\frac{1}{\sin(z)} = \cot(z) + \tan(\frac{z}{2})$? $$\frac{1}{\sin(z)} = \cot (z) + \tan (\tfrac{z}{2})$$
I did this: 
First attempt: $$\displaystyle{\frac{1}{\sin (z)} = \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} = \frac{\cos (z) }{\sin (z)} + \frac{2\sin(\frac{z}{4})\cos(\frac{z}{4})}{\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4})}} = $$
$$\frac{\cos (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))+2\sin z \sin(\frac{z}{4})\cos(\frac{z}{4})}{\sin (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))}$$
Stuck.
Second attempt: 
$$\displaystyle{\frac{1}{\sin z} = \left(\frac{1}{2i}(e^{iz}-e^{-iz})\right)^{-1} = 2i\left(\frac{1}{e^{iz}-e^{-iz}}\right)}$$
Stuck.
Does anybody see a way to continue?
 [/INST] I'll go backwards; I hope you don't mind.
$$\begin{align*}\cot\,z+\tan\frac{z}{2}&=\frac{\cos\,z}{\sin\,z}+\frac{\sin\,z}{1+\cos\,z}\\&=\frac{\sin^2 z+(1+\cos\,z)\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{\cos^2 z+\sin^2 z+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{1+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\csc\,z\end{align*}$$
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