If \( \sec \theta=\sqrt{2} \) and \( \frac{3 \pi}{2}<\theta<2 \pi \), find the value of \( \frac{1+\tan \theta+\operatorname{cosec} \theta}{1+\cot \theta-\operatorname{cosec} \theta} \)

Given that \( \sec \theta = \sqrt{2} \), we can find \( \cos \theta \) as follows: \[ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \] Since \( \frac{3\pi}{2} < \theta < 2\pi \), we know that \( \theta \) is in the fourth quadrant. In the fourth quadrant, \( \cos \theta \) is positive, and thus: \[ \cos \theta = \frac{\sqrt{2}}{2} \] Next, we find \( \sin \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \cos \theta \): \[ \sin^2 \theta + \left(\frac{\sqrt{2}}{2}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{2}{4} = 1 \] \[ \sin^2 \theta + \frac{1}{2} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{2} = \frac{1}{2} \] Taking the square root, we find: \[ \sin \theta = -\frac{\sqrt{2}}{2} \] (since \( \sin \theta \) is negative in the fourth quadrant). Now we can calculate \( \tan \theta \), \( \cot \theta \), and \( \csc \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 \] \[ \cot \theta = \frac{1}{\tan \theta} = -1 \] \[ \csc \theta = \frac{1}{\sin \theta} = -\frac{2}{\sqrt{2}} = -\sqrt{2} \] Now we substitute these values into the expression \( \frac{1+\tan \theta+\csc \theta}{1+\cot \theta-\csc \theta} \): Calculating the numerator: \[ 1 + \tan \theta + \csc \theta = 1 - 1 - \sqrt{2} = 0 - \sqrt{2} = -\sqrt{2} \] Calculating the denominator: \[ 1 + \cot \theta - \csc \theta = 1 - 1 + \sqrt{2} = 0 + \sqrt{2} = \sqrt{2} \] Now we can compute the entire expression: \[ \frac{1+\tan \theta+\csc \theta}{1+\cot \theta-\csc \theta} = \frac{-\sqrt{2}}{\sqrt{2}} = -1 \] Thus, the final answer is: \[ \boxed{-1} \]