Let theta, phi in [0, 2pi] be such that 2 cos | vedclass

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(C) Given $ an (2\pi - heta) > 0$ $\Rightarrow - an heta > 0$ $\Rightarrow an heta < 0$. Also,$-1 < \sin heta < -\frac{\sqrt{3}}{2}$ impli

Vedclass · Accepted Answer

$\frac{4\pi}{3} < \phi < \frac{3\pi}{2}$

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(C) Given $	an (2\pi - 	heta) > 0$ $\Rightarrow -	an 	heta > 0$ $\Rightarrow 	an 	heta Also,$-1 Since $	an 	heta Using the identity $	an \frac{	heta}{2} + \cot \frac{	heta}{2} = \frac{2}{\sin 	heta}$,the equation becomes: $2 \cos 	heta (1 - \sin \phi) = \sin^2 	heta \left(\frac{2}{\sin 	heta}ight) \cos \phi - 1$ $2 \cos 	heta - 2 \cos 	heta \sin \phi = 2 \sin 	heta \cos \phi - 1$ $2 \cos 	heta + 1 = 2(\sin 	heta \cos \phi + \cos 	heta \sin \phi) = 2 \sin(	heta + \phi)$. For $	heta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$2 \cos 	heta + 1 \in (1, 2)$. Thus,$1 This implies $	heta + \phi \in (\frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi)$. For $k=0$,$\phi \in (\frac{\pi}{6} - 	heta, \frac{5\pi}{6} - 	heta)$. Since $	heta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$\phi \in (\frac{\pi}{6} - \frac{5\pi}{3}, \frac{5\pi}{6} - \frac{3\pi}{2}) = (-\frac{3\pi}{2}, -\frac{2\pi}{3})$. Adjusting to $[0, 2\pi]$,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$. For $k=1$,$\phi \in (\frac{13\pi}{6} - 	heta, \frac{17\pi}{6} - 	heta) = (\frac{13\pi}{6} - \frac{5\pi}{3}, \frac{17\pi}{6} - \frac{3\pi}{2}) = (\frac{\pi}{2}, \frac{4\pi}{3})$. Thus,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$. Therefore,$\phi$ cannot satisfy $(A), (C), (D)$.

Let $\theta, \phi \in [0, 2\pi]$ be such that $2 \cos \theta(1-\sin \phi) = \sin^2 \theta \left(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}\right) \cos \phi - 1$,$\tan (2\pi - \theta) > 0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

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