Add or subtract as indicated, and then simplify as much as possible. Leave your answer in terms of sines and cosines only. \( \cos (x)+\frac{1}{\sin (x)}=\square \)

To solve the equation, we start by rewriting the term \(\frac{1}{\sin(x)}\) using sine and cosine. We know that \(\frac{1}{\sin(x)} = \csc(x)\), but as per your request, we want to keep it strictly in terms of sine and cosine. Thus, the expression becomes: \[ \cos(x) + \frac{1}{\sin(x)} = \cos(x) + \frac{\cos(x)}{\sin(x)} \cdot \sin(x) = \cos(x) + \cot(x) \] However, knowing that \(\cot(x) = \frac{\cos(x)}{\sin(x)}\), we can combine the terms into a single fraction. To do this we will multiply \(\cos(x)\) by \(\frac{\sin(x)}{\sin(x)}\): \[ \cos(x) = \frac{\cos(x) \sin(x)}{\sin(x)} \] Now combining the two terms results in: \[ \frac{\cos(x) \sin(x) + \cos(x)}{\sin(x)} = \frac{\cos(x)(\sin(x) + 1)}{\sin(x)} \] Thus we have: \[ \cos(x) + \frac{1}{\sin(x)} = \frac{\cos(x)(\sin(x) + 1)}{\sin(x)} \] And this can be simplified down as needed, but it will remain in terms of sine and cosine!