FRCISE 6 Evaluate, without using a calculator, \( \begin{array}{ll}\text { (a) } \cos 330^{\circ} \cdot \sin 60^{\circ} & \text { (b) } \frac{\sin 135^{\circ}}{\cos 225^{\circ}} \\ \text { (c) } \tan 315^{\circ}-2 \cos 60^{\circ}+\sin 210^{\circ} & \text { (d) } \frac{\sin 570^{\circ}+\cos 240^{\circ}-\tan 135^{\circ}}{\sin 410^{\circ}} \\ \text { (c) } \frac{\tan 330^{\circ}}{\sin 330^{\circ}} & \text { (f) } \frac{\cos 40^{\circ}}{\sin 150^{\circ}-\tan 240^{\circ} \cdot \cos 210^{\circ}} \\ \begin{array}{ll}\text { (g) } & \text { (h) } \tan 120^{\circ} \cdot \cos 210^{\circ}-\sin ^{2} 315^{\circ} \\ \text { (g) } \tan 150^{\circ} \\ \text { (i) } \frac{\sin 300^{\circ}}{\tan 240^{\circ}} & \text { (in } 120^{\circ}\end{array} & \text { (i) } \frac{\tan 315^{\circ}-\cos 1020^{\circ}}{\sin 150^{\circ}+\tan \left(-135^{\circ}\right)} \\ \text { (k) } \frac{\tan 225^{\circ}-\sin x}{\sin x+\sin 270^{\circ}} & \text { (l) } \cos \left(-315^{\circ}\right) \cdot \sin 315^{\circ}-\frac{\cos 20^{\circ}}{\sin 250^{\circ}}\end{array} \)

Sure, let’s dive into evaluating these trigonometric expressions step by step. ### (a) \( \cos 330^{\circ} \cdot \sin 60^{\circ} \) - \(\cos 330^{\circ} = \cos(360^{\circ}-30^{\circ}) = \cos 30^{\circ} = \frac{\sqrt{3}}{2}\) - \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) - So, \(\cos 330^{\circ} \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4}\) ### (b) \( \frac{\sin 135^{\circ}}{\cos 225^{\circ}} \) - \(\sin 135^{\circ} = \sin(180^{\circ}-45^{\circ}) = \sin 45^{\circ} = \frac{\sqrt{2}}{2}\) - \(\cos 225^{\circ} = \cos(180^{\circ}+45^{\circ}) = -\cos 45^{\circ} = -\frac{\sqrt{2}}{2}\) - Thus, \( \frac{\sin 135^{\circ}}{\cos 225^{\circ}} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1\) ### (c) \( \tan 315^{\circ}-2 \cos 60^{\circ}+\sin 210^{\circ} \) - \(\tan 315^{\circ} = -1\) - \(\cos 60^{\circ} = \frac{1}{2}\) so \(2 \cos 60^{\circ} = 1\) - \(\sin 210^{\circ} = -\frac{1}{2}\) - Hence, \(\tan 315^{\circ}-2 \cos 60^{\circ}+\sin 210^{\circ} = -1 - 1 - \frac{1}{2} = -2.5\) ### (d) \( \frac{\sin 570^{\circ}+\cos 240^{\circ}-\tan 135^{\circ}}{\sin 410^{\circ}} \) - First, simplify the angles: - \(\sin 570^{\circ} = \sin(360^{\circ}+210^{\circ}) = \sin 210^{\circ} = -\frac{1}{2}\) - \(\cos 240^{\circ} = -\frac{1}{2}\) - \(\tan 135^{\circ} = -1\) - \(\sin 410^{\circ} = \sin(360^{\circ}+50^{\circ}) = \sin 50^{\circ}\) Putting it all together: - \(-\frac{1}{2} - \frac{1}{2} - (-1) = -\frac{1}{2} - \frac{1}{2} + 1 = 0\) - Since \(\sin 50^{\circ} \neq 0\), \(\frac{0}{\sin 50^{\circ}} = 0\) ### (e) \( \frac{\tan 330^{\circ}}{\sin 330^{\circ}} \) - \(\tan 330^{\circ} = \tan(360^{\circ}-30^{\circ}) = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}\) - \(\sin 330^{\circ} = -\frac{1}{2}\) - So, \(\frac{-\frac{1}{\sqrt{3}}}{-\frac{1}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\) ### (f) \( \frac{\cos 40^{\circ}}{\sin 150^{\circ}-\tan 240^{\circ} \cdot \cos 210^{\circ}} \) - \(\sin 150^{\circ} = \frac{1}{2}\) - \(\tan 240^{\circ} = \frac{1}{\sqrt{3}}\) and \(\cos 210^{\circ} = -\frac{\sqrt{3}}{2}\) - Thus, \(\tan 240^{\circ}