\( \begin{array}{cc}\cot \theta=\frac{3}{4} & \text { and } \pi<\theta<\frac{3 \pi}{2} \\ \sin \theta=\quad & \sin (2 \theta)= \\ \operatorname{coc} \theta=\quad & \cos (2 \theta)= \\ \cos \theta= & \tan (2 \theta)= \\ \sec \theta= & \cos \left(\frac{\theta}{2}\right)= \\ \tan \theta= & \tan \left(\frac{\theta}{2}\right)=\end{array} \)

Given that \( \cot \theta = \frac{3}{4} \) and \( \pi < \theta < \frac{3\pi}{2} \), we can find the values of \( \sin \theta \), \( \cos \theta \), and \( \tan \theta \) first. ### Step 1: Find \( \sin \theta \) and \( \cos \theta \) Since \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can express \( \cos \theta \) and \( \sin \theta \) in terms of a right triangle. Let: - Opposite side = 4 (since \( \sin \theta \) is negative in the third quadrant) - Adjacent side = 3 (since \( \cos \theta \) is also negative in the third quadrant) Using the Pythagorean theorem, we find the hypotenuse \( r \): \[ r = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Now we can find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = \frac{\text{opposite}}{r} = \frac{-4}{5} \quad (\text{negative in the third quadrant}) \] \[ \cos \theta = \frac{\text{adjacent}}{r} = \frac{-3}{5} \quad (\text{negative in the third quadrant}) \] ### Step 2: Find \( \tan \theta \) \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-4/5}{-3/5} = \frac{4}{3} \] ### Step 3: Find \( \sec \theta \) and \( \csc \theta \) \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-3/5} = -\frac{5}{3} \] \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-4/5} = -\frac{5}{4} \] ### Step 4: Find \( \sin(2\theta) \) and \( \cos(2\theta) \) Using the double angle formulas: \[ \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(-\frac{4}{5}\right) \left(-\frac{3}{5}\right) = 2 \cdot \frac{12}{25} = \frac{24}{25} \] \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta = \left(-\frac{3}{5}\right)^2 - \left(-\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25} \] ### Step 5: Find \( \tan(2\theta) \) \[ \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{\frac{24}{25}}{-\frac{7}{25}} = -\frac{24}{7} \] ### Step 6: Find \( \cos\left(\frac{\theta}{2}\right) \) and \( \tan\left(\frac{\theta}{2}\right) \) Using the half-angle formulas: \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \] Since \( \frac{\theta}{2} \) is in the second quadrant (as \( \pi < \theta < \frac{3\pi}{2} \)), \( \cos\left(\frac{\theta}{2}\right) \) is positive. Now, to find \( \tan\left(\frac{\theta}{2}\right) \): \[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin \theta}{1 + \cos \theta} = \frac{-\frac{4}{5}}{1 - \frac{3}{5}} = \frac{-\frac{4}{5}}{\frac{2}{5}} = -2 \] ### Summary of Results \[ \begin{array}{cc} \sin \theta = -\frac{4}{5} & \sin(2\theta) = \frac{24}{25} \\ \csc \theta = -\frac{5}{4} & \cos(2\theta) = -\frac{7}{25} \\ \cos \theta = -\frac{3}{5} & \tan(2\theta) = -\frac{24}{7} \\ \sec \theta = -\frac{5}{3} & \cos\left(\frac{\theta}{2}\right) = \frac{\sqrt{5}}{5} \\ \tan \theta = \frac{4}{3} & \tan\left(\frac{\theta}{2}\right) = -2 \\ \end{array} \]