4. Given: \( p \sin \beta=3 \) and 4.1 In which quadrat 4.2 Show that \( \tan \beta \) 4.3 Determine the \( n t \) 5. Given: \( \cos ^{2} \theta\left(\frac{1}{\sin \theta-1}\right. \) 5.1 For what values 5.2 Prove that \( \cos ^{2} \)

To tackle the first part, we know from \( p \sin \beta = 3 \) that \( \sin \beta \) can take values between -1 and 1. So, we can infer that since \( p \sin \beta \) equals a positive value, \( \beta \) must be in either the first or second quadrant, where sine is positive. Now for the proof of \( \tan \beta \), knowing \( \sin \beta = \frac{3}{p} \) leads to \( \tan \beta = \frac{\sin \beta}{\cos \beta} \). Using Pythagorean identity \( \cos ^2 \beta = 1 - \sin ^2 \beta \), we can derive \( \tan \beta \) in terms of \( p \), culminating in the expression \( \tan \beta = \frac{3}{\sqrt{p^2 - 9}} \), given \( p^2 > 9 \) to maintain the validity of the square root. Shifting gears to the second equation, \( \cos^{2} \theta\left(\frac{1}{\sin \theta-1}\right) \), we need to examine possible values for \( \sin \theta \). Notably, \( \sin \theta \) must be greater than 1 for the function to be defined, which is impossible since \( \sin \theta \) is always between -1 and 1. Thus, there are no valid values of \( \theta \) that satisfy this condition. In proving \( \cos^{2} \), we see that for any \( \theta \) where defined, the maximum value \( \cos^{2} \theta \) can take is 1 (when \( \theta = 0, \pi, \ldots \)). Hence, \( \cos^{2} \theta \) is bounded by 0 and 1, aligning with its definition in the unit circle.