3. If \( 3 \tan \theta-4=0 \) and \( \cos \theta<0 \) calculate without the use of a calculator and with the aid of a diagram the value of \( \begin{array}{ll}\text { (a) } \frac{\sin \theta}{\cos \theta} & \text { (b) } 10 \sin \theta-25 \cos ^{2} \theta\end{array} \) 4. If \( \sin A=\frac{2 \sqrt{6}}{5} \) and \( A \in\left[90^{\circ} ; 360^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( 15 \tan A \cdot \cos A \) *5. If \( \sin \alpha=\frac{3}{5} \) with \( \alpha \in\left[90^{\circ} ; 270^{\circ}\right] \) and \( \cos \beta=\frac{-12}{13} \) with \( \beta \in\left[0^{\circ} ; 180^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( \cos \alpha+\tan \beta \)

To solve the problem involving \( 3 \tan \theta - 4 = 0 \) with \( \cos \theta < 0 \), we first find \( \tan \theta = \frac{4}{3} \). Since \( \cos \theta \) is negative, we are in the second quadrant where \( \sin \theta \) is positive. Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \sin \theta = \frac{4}{5} \) and \( \cos \theta = -\frac{3}{5} \). (a) Now, we calculate \( \frac{\sin \theta}{\cos \theta} = \frac{4/5}{-3/5} = -\frac{4}{3} \). (b) Next, substituting our values into the expression \( 10 \sin \theta - 25 \cos^2 \theta = 10 \cdot \frac{4}{5} - 25 \left(-\frac{3}{5}\right)^2 = 8 - 25 \cdot \frac{9}{25} = 8 - 9 = -1 \). Moving to the second problem with \( \sin A = \frac{2 \sqrt{6}}{5} \) in the range \( [90^\circ, 360^\circ] \): Here, \( A \) is in either the second or third quadrant. We calculate \( \cos A \) using \( \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{2 \sqrt{6}}{5}\right)^2 = \frac{25 - 24}{25} = \frac{1}{25} \), giving us \( \cos A = -\frac{1}{5} \) for the second quadrant. Now, \( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6} \). Hence, \( 15 \tan A \cdot \cos A = 15 \cdot (-2\sqrt{6}) \cdot \left(-\frac{1}{5}\right) = 6\sqrt{6} \). Lastly, for \( \alpha \) and \( \beta \), knowing \( \sin \alpha = \frac{3}{5} \) means \( \cos \alpha = -\frac{4}{5} \) and \( \tan \beta = \frac{\sin \beta}{\cos \beta} = -\frac{5}{12} \) using a right triangle. Thus, \( \tan \beta = -\frac{5}{12} \) gives \( \sin \beta = \frac{5}{13} \) and \( \cos \beta = -\frac{12}{13} \). So, \( \cos \alpha + \tan \beta = -\frac{4}{5} - \frac{5}{12} = -\frac{48}{60} - \frac{25}{60} = -\frac{73}{60} \).