Question 16, 1.3.93 HW Score: 7 Points: Use the trigonometric function values of the quadrantal angles to evaluate the given expression. \[ 3\left(\csc 90^{\circ}\right)^{2}+5\left(\tan 180^{\circ}\right)^{2}+\sec 180^{\circ} \] \[ 3\left(\csc 90^{\circ}\right)^{2}+5\left(\tan 180^{\circ}\right)^{2}+\sec 180^{\circ}= \] \( \square \) (Simplity your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To evaluate the expression: \[ 3\left(\csc 90^{\circ}\right)^{2} + 5\left(\tan 180^{\circ}\right)^{2} + \sec 180^{\circ} \] Let's break it down term by term: 1. **\(\csc 90^{\circ}\):** \[ \csc \theta = \frac{1}{\sin \theta} \] \[ \csc 90^{\circ} = \frac{1}{\sin 90^{\circ}} = \frac{1}{1} = 1 \] \[ \left(\csc 90^{\circ}\right)^2 = 1^2 = 1 \] \[ 3 \times 1 = 3 \] 2. **\(\tan 180^{\circ}\):** \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \tan 180^{\circ} = \frac{\sin 180^{\circ}}{\cos 180^{\circ}} = \frac{0}{-1} = 0 \] \[ \left(\tan 180^{\circ}\right)^2 = 0^2 = 0 \] \[ 5 \times 0 = 0 \] 3. **\(\sec 180^{\circ}\):** \[ \sec \theta = \frac{1}{\cos \theta} \] \[ \sec 180^{\circ} = \frac{1}{\cos 180^{\circ}} = \frac{1}{-1} = -1 \] Now, combine all the results: \[ 3 + 0 - 1 = 2 \] **Answer:** \(2\)