i) \( \frac{\operatorname{sen}\left(\beta-\frac{3 \pi}{2}\right)}{\sec \beta}+\frac{\cos \left(\frac{\pi}{2}-\beta\right)}{\csc \beta}=1 \)

To solve the equation \[ \frac{\operatorname{sen}\left(\beta-\frac{3 \pi}{2}\right)}{\sec \beta}+\frac{\cos \left(\frac{\pi}{2}-\beta\right)}{\csc \beta}=1, \] we start by simplifying each term. First, we know that: \[ \operatorname{sen}\left(\beta - \frac{3\pi}{2}\right) = \sin\left(\beta - \frac{3\pi}{2}\right) = \sin\left(\beta + \frac{3\pi}{2}\right) = -\cos(\beta). \] Then, \(\sec \beta = \frac{1}{\cos \beta}\). Thus, we can rewrite the first term: \[ \frac{\operatorname{sen}\left(\beta-\frac{3 \pi}{2}\right)}{\sec \beta} = \frac{-\cos \beta}{\frac{1}{\cos \beta}} = -\cos^2 \beta. \] Next, for the second term, we can simplify it as follows. We know that: \[ \cos\left(\frac{\pi}{2} - \beta\right) = \sin \beta, \] and since \(\csc \beta = \frac{1}{\sin \beta}\), we have: \[ \frac{\cos\left(\frac{\pi}{2}-\beta\right)}{\csc \beta} = \frac{\sin \beta}{\frac{1}{\sin \beta}} = \sin^2 \beta. \] Now, substituting these back into the original equation gives us: \[ -\cos^2 \beta + \sin^2 \beta = 1. \] Next, we can use the identity \(\sin^2 \beta + \cos^2 \beta = 1\). This can be rearranged as \(\sin^2 \beta = 1 - \cos^2 \beta\). Thus, we can express the equation as: \[ - \cos^2 \beta + (1 - \cos^2 \beta) = 1. \] This simplifies to: \[ 1 - 2\cos^2 \beta = 1, \] leading to: \[ -2\cos^2 \beta = 0. \] Thus, \(\cos^2 \beta = 0\), which means: \[ \cos \beta = 0. \] The solutions for \(\beta\) where \(\cos \beta = 0\) are: \[ \beta = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z}. \] In conclusion, the general solutions to the equation are: \[ \beta = \frac{\pi}{2} + k\pi \text{ for } k \text{ being any integer}. \]