A expressão $\dfrac{\sin \theta}{1 + \cos \theta}$, $0 < \theta < \pi$, idêntica a:


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Augusto Massayoshi 29/03/2022 03:17
$$ E = \frac{\sin \theta}{1 + \cos \theta} $$ — \begin{equation} \tag{1} \cos \theta = 2\cos^2 \left( \frac{\theta}{2} \right) -1 \end{equation} \begin{equation} \tag{2} \sin \theta = 2 \sin \left(\frac{\theta}{2} \right) \cos \left(\frac{\theta}{2} \right) \end{equation} Aplicando $(1)$ e $(2)$ em $E$: \begin{align} E & = \frac{ 2 \sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)}{ 1 + 2\cos^2 \left( \frac{\theta}{2} \right) -1 } \\[5pt] & = \displaystyle \frac{ 2 \sin \left(\frac{\theta}{2} \right) \cos \left(\frac{\theta}{2} \right) }{ 2\cos^2 \left( \frac{\theta}{2}\right) } \\[5pt] & = \displaystyle \frac{\sin \left(\frac{\theta}{2}\right) }{ \cos \left( \frac{\theta}{2} \right) } \\[5pt] & = \tan \left( \frac{\theta}{2} \right) \end{align} Alternativa correta: $ \boxed{\mathrm{D}}$ $$ \boxed{E = \tan \left( \frac{\theta}{2} \right) } $$
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