1. Given: $\sin \alpha = -\frac{1}{3}$ and $\cos \beta = -\frac{1}{2}$. We need to find $\sin(\alpha + \beta) \sin(\alpha - \beta)$. 2. Use the formulas: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$ 3. Multiply these two: $$\sin(\alpha + \beta) \sin(\alpha - \beta) = (\sin \alpha \cos \beta)^2 - (\cos \alpha \sin \beta)^2$$ 4. We know $\sin \alpha = -\frac{1}{3}$ and $\cos \beta = -\frac{1}{2}$. Find $\cos \alpha$ and $\sin \beta$ using Pythagorean identity: $$\cos \alpha = \pm \sqrt{1 - \sin^2 \alpha} = \pm \sqrt{1 - \left(-\frac{1}{3}\right)^2} = \pm \sqrt{1 - \frac{1}{9}} = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}$$ $$\sin \beta = \pm \sqrt{1 - \cos^2 \beta} = \pm \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \pm \sqrt{1 - \frac{1}{4}} = \pm \frac{\sqrt{3}}{2}$$ 5. Since $\sin \alpha$ is negative, $\alpha$ is in either third or fourth quadrant, so $\cos \alpha$ is positive in third quadrant and negative in fourth. Without quadrant info, assume $\cos \alpha = \frac{2\sqrt{2}}{3}$. Similarly, $\cos \beta$ is negative, so $\beta$ is in second or third quadrant. $\sin \beta$ is positive in second quadrant and negative in third. Assume $\sin \beta = \frac{\sqrt{3}}{2}$. 6. Calculate: $$(\sin \alpha \cos \beta)^2 = \left(-\frac{1}{3} \times -\frac{1}{2}\right)^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36}$$ $$(\cos \alpha \sin \beta)^2 = \left(\frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{2}\right)^2 = \left(\frac{2\sqrt{2} \times \sqrt{3}}{3 \times 2}\right)^2 = \left(\frac{\sqrt{6}}{3}\right)^2 = \frac{6}{9} = \frac{2}{3}$$ 7. Substitute: $$\sin(\alpha + \beta) \sin(\alpha - \beta) = \frac{1}{36} - \frac{2}{3} = \frac{1}{36} - \frac{24}{36} = -\frac{23}{36}$$ --- 8. Given: $\sin x \cos y = -\frac{1}{3}$ and $\cos x \sin y = \frac{2}{3}$. Find $\cot(x - y)$. 9. Use formula: $$\cot(x - y) = \frac{\cos(x - y)}{\sin(x - y)} = \frac{\cos x \cos y + \sin x \sin y}{\sin x \cos y - \cos x \sin y}$$ 10. Calculate numerator: $$\cos x \cos y + \sin x \sin y = ?$$ 11. Calculate denominator: $$\sin x \cos y - \cos x \sin y = -\frac{1}{3} - \frac{2}{3} = -1$$ 12. Use identity: $$(\sin x \cos y)^2 + (\cos x \sin y)^2 = \left(-\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}$$ 13. Since $\sin^2 x + \cos^2 x = 1$ and $\sin^2 y + \cos^2 y = 1$, we can find $\cos x \cos y + \sin x \sin y$ using: $$\cos(x - y) = \cos x \cos y + \sin x \sin y$$ 14. Use the identity: $$(\cos x \cos y + \sin x \sin y)^2 + (\sin x \cos y - \cos x \sin y)^2 = (\cos^2 x + \sin^2 x)(\cos^2 y + \sin^2 y) = 1$$ 15. Substitute known values: $$(\cos x \cos y + \sin x \sin y)^2 + (-1)^2 = 1$$ $$ (\cos x \cos y + \sin x \sin y)^2 = 0$$ 16. So: $$\cos x \cos y + \sin x \sin y = 0$$ 17. Therefore: $$\cot(x - y) = \frac{0}{-1} = 0$$ --- 18. Calculate $(\tan 60^\circ \cos 15^\circ - \sin 15^\circ) \times 7\sqrt{2}$. 19. Use known values: $$\tan 60^\circ = \sqrt{3}, \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$$ 20. Calculate inside parentheses: $$\sqrt{3} \times \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{3}(\sqrt{6} + \sqrt{2}) - (\sqrt{6} - \sqrt{2})}{4}$$ 21. Simplify numerator: $$\sqrt{3} \times \sqrt{6} = \sqrt{18} = 3\sqrt{2}$$ $$\sqrt{3} \times \sqrt{2} = \sqrt{6}$$ So numerator: $$3\sqrt{2} + \sqrt{6} - \sqrt{6} + \sqrt{2} = 3\sqrt{2} + \sqrt{2} = 4\sqrt{2}$$ 22. So expression inside parentheses: $$\frac{4\sqrt{2}}{4} = \sqrt{2}$$ 23. Multiply by $7\sqrt{2}$: $$\sqrt{2} \times 7\sqrt{2} = 7 \times 2 = 14$$ --- 24. Calculate $4 \cos 20^\circ - \sqrt{3} \cot 20^\circ$. 25. Use $\cot 20^\circ = \frac{\cos 20^\circ}{\sin 20^\circ}$. 26. Expression: $$4 \cos 20^\circ - \sqrt{3} \frac{\cos 20^\circ}{\sin 20^\circ} = \cos 20^\circ \left(4 - \frac{\sqrt{3}}{\sin 20^\circ}\right)$$ 27. Use approximate values: $$\cos 20^\circ \approx 0.9397, \sin 20^\circ \approx 0.3420$$ 28. Calculate: $$4 - \frac{\sqrt{3}}{0.3420} = 4 - \frac{1.732}{0.3420} = 4 - 5.067 = -1.067$$ 29. Multiply: $$0.9397 \times -1.067 \approx -1$$ --- 30. Calculate $$\left(\frac{\tan^2 \frac{7\pi}{24} - \tan^2 \frac{\pi}{24}}{1 - \tan^2 \frac{7\pi}{24} \tan^2 \frac{\pi}{24}}\right)^2$$. 31. Use formula for tangent difference: $$\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$$ 32. The expression inside parentheses is: $$\tan^2(a - b)$$ where $a = \frac{7\pi}{24}$ and $b = \frac{\pi}{24}$. 33. So the expression is $(\tan^2(a - b))^2 = \tan^4(a - b)$. 34. Calculate $a - b = \frac{7\pi}{24} - \frac{\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$. 35. So expression is: $$\tan^4 \frac{\pi}{4} = 1^4 = 1$$ --- Final answers: 1) $-\frac{23}{36}$ 2) $0$ 3) $14$ 4) $-1$ 5) $1$