Which of the following is equal to $\dfrac{\sin \theta}{1 + \cos \theta}$. Which of the following is equal to $\dfrac{\sin \theta}{1 + \cos \theta}$. A. $\dfrac{\sin \theta}{2 \sin 2\theta}$ C. $\dfrac{\sin \theta}{1 + \sin \theta}$ B. $\dfrac{\cos \theta}{1 - \sin \theta}$ D. $\dfrac{1 - \cos \theta}{\sin \theta}$ Log in or register to post comments Solution Click here to… Jhun Vert Fri, 06/07/2024 - 21:09 Solution Click here to expand or collapse this section $\begin{align} \dfrac{\sin \theta}{1 + \cos \theta} & = \dfrac{\sin \theta}{1 + \cos \theta} \times \dfrac{1 - \cos \theta}{1 - \cos \theta} \\ \\ & = \dfrac{\sin \theta (1 - \cos \theta)}{1 - \cos^2 \theta} \\ \\ & = \dfrac{\sin \theta (1 - \cos \theta)}{\sin^2 \theta} \\ \\ & = \dfrac{1 - \cos \theta}{\sin \theta} \end{align}$ Log in or register to post comments
Solution Click here to… Jhun Vert Fri, 06/07/2024 - 21:09 Solution Click here to expand or collapse this section $\begin{align} \dfrac{\sin \theta}{1 + \cos \theta} & = \dfrac{\sin \theta}{1 + \cos \theta} \times \dfrac{1 - \cos \theta}{1 - \cos \theta} \\ \\ & = \dfrac{\sin \theta (1 - \cos \theta)}{1 - \cos^2 \theta} \\ \\ & = \dfrac{\sin \theta (1 - \cos \theta)}{\sin^2 \theta} \\ \\ & = \dfrac{1 - \cos \theta}{\sin \theta} \end{align}$ Log in or register to post comments
Solution Click here to…