5. Determine the exact value for the following: (5 marks) \[ \sin \frac{11 \pi}{6} \sec \frac{2 \pi}{3}+\tan \frac{5 \pi}{4} \]

To determine the exact value of the expression: \[ \sin \frac{11\pi}{6} \sec \frac{2\pi}{3} + \tan \frac{5\pi}{4} \] we will evaluate each trigonometric function separately. 1. **Evaluating \(\sin \frac{11\pi}{6}\):** - The angle \(\frac{11\pi}{6}\) is in the **4th quadrant** where sine is negative. - The reference angle is \(2\pi - \frac{11\pi}{6} = \frac{\pi}{6}\). - \(\sin \frac{\pi}{6} = \frac{1}{2}\), so: \[ \sin \frac{11\pi}{6} = -\frac{1}{2} \] 2. **Evaluating \(\sec \frac{2\pi}{3}\):** - Recall that \(\sec \theta = \frac{1}{\cos \theta}\). - The angle \(\frac{2\pi}{3}\) is in the **2nd quadrant** where cosine is negative. - The reference angle is \(\pi - \frac{2\pi}{3} = \frac{\pi}{3}\). - \(\cos \frac{\pi}{3} = \frac{1}{2}\), so: \[ \cos \frac{2\pi}{3} = -\frac{1}{2} \] \[ \sec \frac{2\pi}{3} = \frac{1}{\cos \frac{2\pi}{3}} = \frac{1}{-\frac{1}{2}} = -2 \] 3. **Evaluating \(\tan \frac{5\pi}{4}\):** - The angle \(\frac{5\pi}{4}\) is in the **3rd quadrant** where tangent is positive. - The reference angle is \(\frac{5\pi}{4} - \pi = \frac{\pi}{4}\). - \(\tan \frac{\pi}{4} = 1\), so: \[ \tan \frac{5\pi}{4} = 1 \] 4. **Combining the results:** \[ \sin \frac{11\pi}{6} \cdot \sec \frac{2\pi}{3} + \tan \frac{5\pi}{4} = \left(-\frac{1}{2}\right) \cdot (-2) + 1 = 1 + 1 = 2 \] **Final Answer:** \[ 2 \]