EXERCISE 1 (a) If \( -5 \sin \theta=4 \) and \( \theta \in\left(90^{\circ} ; 270^{\circ}\right) \), determine without the use of a calculator and with the aid of a diagram the value of: \( \begin{array}{lll}\text { (1) } 5 \cos \theta-3 \tan \theta & \text { (2) } \frac{4}{\sin \theta}-\frac{3}{\cos \theta}\end{array} \)

To solve \( -5 \sin \theta = 4 \), we first get \( \sin \theta = -\frac{4}{5} \). Given that \( \theta \) is in the range \( (90^{\circ}, 270^{\circ}) \), this means \(\theta\) is in the third quadrant where sine is negative. Next, we can find \( \cos \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \left(-\frac{4}{5}\right)^2 + \cos^2 \theta = 1 \implies \frac{16}{25} + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{16}{25} = \frac{9}{25} \implies \cos \theta = -\frac{3}{5}. \] Now we compute both expressions. 1) Calculating \( 5 \cos \theta - 3 \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}. \] Thus, \[ 5 \cos \theta - 3 \tan \theta = 5\left(-\frac{3}{5}\right) - 3\left(\frac{4}{3}\right) = -3 - 4 = -7. \] 2) For \( \frac{4}{\sin \theta} - \frac{3}{\cos \theta} \): \[ \frac{4}{\sin \theta} = \frac{4}{-\frac{4}{5}} = -5, \quad \frac{3}{\cos \theta} = \frac{3}{-\frac{3}{5}} = -5. \] Thus, \[ \frac{4}{\sin \theta} - \frac{3}{\cos \theta} = -5 - (-5) = 0. \] So, the final answers are: 1) \( -7 \) 2) \( 0 \)