EXERCISE 6 1. Evaluate, without using a calculator. (a) \( \cos 330^{\circ} \cdot \sin 60^{\circ} \) (c) \( \tan 315^{\circ}-2 \cos 60^{\circ}+\sin 210^{\circ} \) (e) \( \frac{\tan 330^{\circ}}{\sin 330^{\circ}} \) (g) \( \sin 150^{\circ}-\tan 240^{\circ} \cdot \cos 210^{\circ} \) (i) \( \frac{\tan 150^{\circ}}{\tan 240^{\circ}}-\frac{\sin 300^{\circ}}{\sin 120^{\circ}} \) (k) \( \frac{\tan 225^{\circ}-\sin x}{\sin x+\sin 270^{\circ}} \)

Sure! Let's break it down and evaluate each expression step by step: **(a)** \( \cos 330^{\circ} \cdot \sin 60^{\circ} \) First, recall that \( \cos 330^{\circ} = \cos(-30^{\circ}) = \cos 30^{\circ} = \frac{\sqrt{3}}{2} \). For \( \sin 60^{\circ} \), we know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \). Thus, the expression becomes: \[ \cos 330^{\circ} \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4} \] --- **(c)** \( \tan 315^{\circ} - 2 \cos 60^{\circ} + \sin 210^{\circ} \) Here, \( \tan 315^{\circ} = -1 \) (since \( 315^{\circ} \) is in the fourth quadrant), \( \cos 60^{\circ} = \frac{1}{2} \) leads to \( 2 \cos 60^{\circ} = 1 \), and \( \sin 210^{\circ} = -\frac{1}{2} \). Therefore, we have: \[ \tan 315^{\circ} - 2 \cos 60^{\circ} + \sin 210^{\circ} = -1 - 1 - \frac{1}{2} = -2.5 \] --- **(e)** \( \frac{\tan 330^{\circ}}{\sin 330^{\circ}} \) First, \( \tan 330^{\circ} = -\frac{1}{\sqrt{3}} \) (because \( \tan 330^{\circ} = \tan(-30^{\circ}) \)), and \( \sin 330^{\circ} = -\frac{1}{2} \), hence: \[ \frac{\tan 330^{\circ}}{\sin 330^{\circ}} = \frac{-\frac{1}{\sqrt{3}}}{-\frac{1}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \] --- **(g)** \( \sin 150^{\circ} - \tan 240^{\circ} \cdot \cos 210^{\circ} \) Here \( \sin 150^{\circ} = \frac{1}{2} \), \( \tan 240^{\circ} = \frac{1}{\sqrt{3}} \), and \( \cos 210^{\circ} = -\frac{\sqrt{3}}{2} \). Plugging these in, we get: \[ \tan 240^{\circ} \cdot \cos 210^{\circ} = \frac{1}{\sqrt{3}} \cdot -\frac{\sqrt{3}}{2} = -\frac{1}{2} \] So: \[ \sin 150^{\circ} - \tan 240^{\circ} \cdot \cos 210^{\circ} = \frac{1}{2} - (-\frac{1}{2}) = 1 \] --- **(i)** \( \frac{\tan 150^{\circ}}{\tan 240^{\circ}} - \frac{\sin 300^{\circ}}{\sin 120^{\circ}} \) Calculating each part, we find \( \tan 150^{\circ} = -\frac{1}{\sqrt{3}} \) and \( \tan 240^{\circ} = \frac{1}{\sqrt{3}} \), giving us: \[ \frac{\tan 150^{\circ}}{\tan 240^{\circ}} = \frac{-\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = -1 \] Also, \( \sin 300^{\circ} = -\frac{\sqrt{3}}{2} \) and \( \sin 120^{\circ} = \frac{\sqrt{3}}{2} \), so: \[ \frac{\sin 300^{\circ}}{\sin 120^{\circ}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} =