Trigonometric Identify: $(1+\cos{x})/\sin{x}=\cot{(x/2)}$

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I am having trouble with this. I have tried many times already. I just can not figure out how to deal with the $x/2$.
$$\frac{1+\cos{x}}{\sin{x}}=\cot{\frac{x}{2}}$$

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4 Answers

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Notice $\cos x = \cos^2 (x/2) - \sin^2 (x/2) $ and $\sin x = 2 \sin (x/2) \cos (x/2) $. Thus,

$$ \frac{ 1 + \cos x }{\sin x } = \frac{ 1 + \cos^2 (x/2) - \sin^2 (x/2) }{2 \sin (x/2) \cos(x/2)} = \frac{2 \cos^2 (x/2) }{2 \sin (x/2) \cos(x/2)}= \frac{ \cos (x/2) }{\sin (x/2)} = \cot(x/2)$$

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$$ \begin{align} \cot(x/2) &=\frac{\cos(x/2)}{\sin(x/2)}\\ &=\frac{2\cos^2(x/2)}{\sin(x)}\\ &=\frac{\cos(x)+1}{\sin(x)}\\ \end{align} $$

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$(\frac{1+\cos{x}}{\sin{x}})^2 = \frac{(1+\cos{x})^2}{1-\cos ^2 x}$

$= \frac{1+\cos x}{1-\cos x} = \frac{\frac{1+\cos x}{2}}{\frac{1-\cos x}{2}}$

$= \frac{\cos^2(x/2)}{\sin^2(x/2)} = \cot^2(\frac{x}{2})$

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$\dfrac{1+cos\,x}{sin\,x}=cot\dfrac{x}{2}$ $$1+cos\,x=2\,cos^2\frac{x}{2}$$ $$sin\,x=2\,sin\frac{x}{2}cos\frac{x}{2}$$ $\Rightarrow \dfrac{2\,cos^2\frac{x}{2}}{2\,sin\frac{x}{2}cos\frac{x}{2}}=\dfrac{cos\frac{x}{2}}{sin\frac{x}{2}}$ $$\dfrac{cos\,x}{sin\,x}=cot\,x$$ $\Rightarrow\dfrac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\dfrac{x}{2}$

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