# 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?

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For your first strategy: You want to compare arguments $z$ and $z/2$, so you should halve $z$ with the formula, not $z/2$. For your second strategy: This is fine, now replace the right side also by exponentials and clear denominators. People post shorter solutions, but you should reflect that both of your ideas actually work fine. – Phira Oct 31 '11 at 23:15

$$\frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} =\frac{\cos (z)\cos (\frac{z}{2})+ \sin(z)\sin (\frac{z}{2}) }{\sin (z)\cos (\frac{z}{2})} =\frac{\cos (z-\frac{z}{2})}{\sin (z)\cos (\frac{z}{2})}$$

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Rather elegant :-) – joriki Oct 31 '11 at 22:35
$\displaystyle{ \frac{cos(-\frac{z}{2})}{sin(z) cos(\frac{z}{2}} = \frac{cos(\frac{z}{2})}{sin(z)cos(\frac{z}{2}} = \frac{1}{sin(z)} }$ – VVV Oct 31 '11 at 22:44

Let $w = \frac{z}{2}$. Then $$\cot(2w) + \tan(w) = \frac{\cos^2(w)-\sin^2(w)}{2 \sin(w) \cos(w)} + \frac{\sin(w)}{\cos(w)} = \frac{1}{\cos(w)} \left( \frac{\cos^2(w)-\sin^2(w) + 2 \sin^2(w)}{2 \sin(w)} \right)$$ The numerator becomes 1, and we arrive at the result $\frac{1}{2 \sin(w) \cos(w)} = \frac{1}{\cos(2w)} = \frac{1}{\cos(z)}$.

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Start out with $$\frac{1-\cos(z)}{\sin(z)}=\frac{2\sin^2(\tfrac{z}{2})}{2\sin(\tfrac{z}{2})\cos(\tfrac{z}{2})}=\tan(\tfrac{z}{2})\tag{1}$$ and add $\cot(z)$ to both sides: $$\frac{1}{\sin(z)}=\cot(z)+\tan(\tfrac{z}{2})\tag{2}$$

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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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