The point \( P(x, y) \) is on the terminal ray of angle \( \theta \). If \( \theta \) is between \( \pi \) radians and \( \frac{3 \pi}{2} \) radians and \( \csc \theta=-\frac{5}{2} \), what are the coordinates of \( P(x, y) \) ? \( P(-\sqrt{21},-2) \) \( P(\sqrt{21},-2) \) \( P(-2, \sqrt{21}) \) \( P(-2,-\sqrt{21}) \)

To find the coordinates of point \( P(x, y) \), we start with the given information about \( \csc \theta \). We know that \( \csc \theta = \frac{1}{\sin \theta} \), and since \( \csc \theta = -\frac{5}{2} \), we have: \[ \sin \theta = -\frac{2}{5} \] Since \( \theta \) is between \( \pi \) radians and \( \frac{3\pi}{2} \) radians, this places \( \theta \) in the third quadrant, where both sine and cosine are negative. We will use the identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for \( \sin \theta \): \[ \left(-\frac{2}{5}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{4}{25} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25} \] Taking the square root, we find \( \cos \theta \): \[ \cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5} \] Now that we have both sine and cosine values, we can express the coordinates of point \( P \) as follows, where the radius \( r \) can be taken as 5 (as it relates to \( c \) where \( r = \frac{1}{\csc \theta} \)): \[ x = r \cos \theta = 5 \left(-\frac{\sqrt{21}}{5}\right) = -\sqrt{21} \] \[ y = r \sin \theta = 5 \left(-\frac{2}{5}\right) = -2 \] Thus, the coordinates of point \( P(x, y) \) are: \[ P(-\sqrt{21}, -2) \] So the correct answer is: \( P(-\sqrt{21},-2) \)