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I need to show that $\frac{\sin\theta}{1+\cos\theta}=\tan\left({\frac{\theta}{2}}\right)$ using the A-level MF26 identities.

I have looked at converting $1+\cos\theta$ to $2\cos^2\left({\frac{\theta}{2}}\right)$ but am still a bit stuck. Would welcome any suggestions.

HVW
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    I don't know MF26, but here is an elementary proof. In the duplicate below, also $\tan \left ( \frac{\alpha}{2} \right ) = \frac{\sin \alpha}{1 + \cos \alpha}$ is shown with the duplication formula. – Dietrich Burde May 21 '23 at 09:11

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Using the double-angle identities \begin{align*} \sin(2x) & = 2\sin x\cos x\\ \cos(2x) & = 2\cos^2x - 1 \end{align*} with $x = \frac{\theta}{2}$ yields $$\frac{\sin\theta}{1 + \cos\theta} = \frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{1 + 2\cos^2\left(\frac{\theta}{2}\right) - 1} = \frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right)$$ provided that $\theta \neq (2k + 1)\pi, k \in \mathbb{Z}$.

N. F. Taussig
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