1. The problem is to find the exact value of $\cot\left(-\frac{5\pi}{12}\right)$. 2. Recall that $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and that cotangent is an odd function: $\cot(-\theta) = -\cot \theta$. 3. So, $\cot\left(-\frac{5\pi}{12}\right) = -\cot\left(\frac{5\pi}{12}\right)$. 4. Express $\frac{5\pi}{12}$ as $\frac{\pi}{3} + \frac{\pi}{4}$, angles with known sine and cosine values. 5. Use the cotangent addition formula: $$\cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$$ 6. Compute $\cot \frac{\pi}{3} = \frac{\cos \frac{\pi}{3}}{\sin \frac{\pi}{3}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$. 7. Compute $\cot \frac{\pi}{4} = 1$. 8. Substitute in formula: $$\cot\left(\frac{\pi}{3}+\frac{\pi}{4}\right) = \frac{\frac{1}{\sqrt{3}} \cdot 1 -1}{\frac{1}{\sqrt{3}} +1} = \frac{\frac{1}{\sqrt{3}} -1}{\frac{1}{\sqrt{3}} +1}$$ 9. Multiply numerator and denominator by $\sqrt{3}$ to rationalize: $$= \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$$ 10. Rationalize denominator: $$= \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{1 - 3} = \frac{1 - 2\sqrt{3} + 3}{-2} = \frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3}$$ 11. Recall step 3, the original value is: $$\cot\left(-\frac{5\pi}{12}\right) = -\cot\left(\frac{5\pi}{12}\right) = -(-2 + \sqrt{3}) = 2 - \sqrt{3}$$ --- 12. The problem is to find the exact value of $\sin 105^\circ - \cos 15^\circ$. 13. Use angle sum and difference identities: $$\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$$ 14. Calculate each term: $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$. 15. So, $$\sin 105^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 16. For $\cos 15^\circ = \cos(45^\circ - 30^\circ)$: $$\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$$ 17. Calculate: $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$. 18. Substitute: $$\cos 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 19. Thus, $$\sin 105^\circ - \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{6} + \sqrt{2}}{4} = 0$$ --- 20. The problem is to find the exact value of $\tan 1875^\circ$. 21. Reduce the angle modulo $360^\circ$ since tangent has period $180^\circ$: $$1875^\circ - 10 \times 180^\circ = 1875^\circ - 1800^\circ = 75^\circ$$ 22. So, $$\tan 1875^\circ = \tan 75^\circ$$ 23. Recall $\tan (45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$ 24. Rationalize denominator: $$= \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$ 25. Therefore, $$\tan 1875^\circ = 2 + \sqrt{3}$$ --- 26. The problem is: Let $\alpha, \beta$ be acute angles such that $\cot \alpha = 7$ and $\csc \beta = \sqrt{10}$. Find $\cos(\alpha+\beta)$. 27. From $\cot \alpha = 7$, $\tan \alpha = \frac{1}{7}$, opposite side $=1$, adjacent side $=7$, hypotenuse $= \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2}$. 28. So, $$\sin \alpha = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}$$ $$\cos \alpha = \frac{7}{5\sqrt{2}} = \frac{7\sqrt{2}}{10}$$ 29. Given $\csc \beta = \sqrt{10}$, so $\sin \beta = \frac{1}{\sqrt{10}}$. 30. Hypotenuse = 1, opposite = $\sin \beta = \frac{1}{\sqrt{10}}$, adjacent side = $\sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$. 31. Thus, $$\cos \beta = \frac{3}{\sqrt{10}}$$ 32. Use the cosine addition formula: $$\cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ 33. Substitute values: $$= \frac{7\sqrt{2}}{10} \cdot \frac{3}{\sqrt{10}} - \frac{\sqrt{2}}{10} \cdot \frac{1}{\sqrt{10}} = \frac{21 \sqrt{2}}{10 \sqrt{10}} - \frac{\sqrt{2}}{10 \sqrt{10}} = \frac{20 \sqrt{2}}{10 \sqrt{10}} = \frac{2 \sqrt{2}}{\sqrt{10}}$$ 34. Simplify $\frac{2 \sqrt{2}}{\sqrt{10}} = 2 \sqrt{\frac{2}{10}} = 2 \sqrt{\frac{1}{5}} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$. --- 35. The problem is: Given $\sin \alpha = -\frac{8}{17}$ and $\sin \beta = -\frac{1}{2}$ with $\alpha, \beta$ in quadrant IV, find $\cos(\alpha + \beta)$. 36. In quadrant IV, $\sin$ is negative and $\cos$ is positive. 37. Find $\cos \alpha$: $$\cos \alpha = +\sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(-\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$$ 38. Find $\cos \beta$: $$\cos \beta = +\sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$$ 39. Use cosine addition formula: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ 40. Substitute: $$= \frac{15}{17} \cdot \frac{\sqrt{3}}{2} - \left(-\frac{8}{17}\right) \cdot \left(-\frac{1}{2}\right) = \frac{15 \sqrt{3}}{34} - \frac{8}{34} = \frac{15 \sqrt{3} - 8}{34}$$ --- 41. The problem: If $3 \sin x = 2$, find $\sin(x - \pi) + \sin(x + \pi)$. 42. Solve for $\sin x$: $$\sin x = \frac{2}{3}$$ 43. Use sine addition formulas: $$\sin(x - \pi) = \sin x \cos \pi - \cos x \sin \pi = \sin x \cdot (-1) - \cos x \cdot 0 = -\sin x$$ 44. Similarly, $$\sin(x + \pi) = \sin x \cos \pi + \cos x \sin \pi = -\sin x$$ 45. Sum: $$\sin(x - \pi) + \sin(x + \pi) = -\sin x - \sin x = -2 \sin x = -2 \cdot \frac{2}{3} = -\frac{4}{3}$$ 46. Note: This result is outside the usual sine range $[-1,1]$ implying the problem might be conceptual, but based on calculations, it is correct. --- 47. The problem: Simplify $\cos(x + \frac{\pi}{2}) + \cos(\frac{\pi}{2} - x)$. 48. Use the identity: $$\cos \left(x + \frac{\pi}{2}\right) = -\sin x$$ 49. Also, $$\cos \left(\frac{\pi}{2} - x\right) = \sin x$$ 50. Sum: $$-\sin x + \sin x = 0$$ --- 51. The problem: Given $\sin A = \frac{4}{5}$ with $\frac{\pi}{2} \le A \le \pi$ (Quadrant II or III) and $\cos B = \frac{4}{5}$ with $B$ not in QI. Find: (a) $\sin(A-B)$, (b) $\cos(A-B)$, (c) $\tan(A-B)$, and determine the quadrant of $A-B$. 52. Since $\sin A = 4/5$ and $A$ in $[\pi/2, \pi]$, $A$ is in quadrant II. There, $\sin$ positive, $\cos$ negative. 53. Find $\cos A$: $$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$ 54. Given $\cos B = 4/5$ and $B$ not in QI. Since cosine is positive, $B$ is in quadrant IV. There, $\cos >0$, $\sin <0$. 55. Find $\sin B$: $$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\frac{3}{5}$$ 56. Use sine difference formula: $$\sin(A-B) = \sin A \cos B - \cos A \sin B = \frac{4}{5} \cdot \frac{4}{5} - \left(-\frac{3}{5}\right) \cdot \left(-\frac{3}{5}\right) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$ 57. Use cosine difference formula: $$\cos(A-B) = \cos A \cos B + \sin A \sin B = \left(-\frac{3}{5}\right) \cdot \frac{4}{5} + \frac{4}{5} \cdot \left(-\frac{3}{5}\right) = -\frac{12}{25} - \frac{12}{25} = -\frac{24}{25}$$ 58. Find tangent difference: $$\tan(A-B) = \frac{\sin(A-B)}{\cos(A-B)} = \frac{7/25}{-24/25} = -\frac{7}{24}$$ 59. Determine quadrant of $A-B$: Since $\sin(A-B) > 0$ and $\cos(A-B) < 0$, $A-B$ is in quadrant II. --- 60. The problem: Given $\csc A = \sqrt{3}$, $A$ in QI, and $\sec B = \sqrt{2}$, $\sin B < 0$, find: (a) $\sin(A-B)$, (b) $\cos(A-B)$, (c) $\tan(A-B)$, and determine the quadrant of $A-B$. 61. From $\csc A = \sqrt{3}$, $\sin A = 1/\sqrt{3} = \frac{\sqrt{3}}{3}$. 62. Since $A$ in QI, $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$. 63. From $\sec B = \sqrt{2}$, $\cos B = 1/\sqrt{2} = \frac{\sqrt{2}}{2}$. 64. Since $\sin B < 0$ and $\cos B > 0$, $B$ is in quadrant IV. 65. Find $\sin B$: $$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \frac{1}{2}} = -\frac{\sqrt{2}}{2}$$ 66. Compute $\sin(A-B)$: $$= \sin A \cos B - \cos A \sin B = \frac{\sqrt{3}}{3} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{3} \cdot \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{6}}{6} + \frac{\sqrt{12}}{6} = \frac{\sqrt{6}}{6} + \frac{2\sqrt{3}}{6} = \frac{\sqrt{6} + 2\sqrt{3}}{6}$$ 67. Compute $\cos(A-B)$: $$= \cos A \cos B + \sin A \sin B = \frac{\sqrt{6}}{3} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} \cdot \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{12}}{6} - \frac{\sqrt{6}}{6} = \frac{2 \sqrt{3} - \sqrt{6}}{6}$$ 68. Compute $\tan(A-B)$: $$= \frac{\sin(A-B)}{\cos(A-B)} = \frac{\frac{\sqrt{6} + 2\sqrt{3}}{6}}{\frac{2 \sqrt{3} - \sqrt{6}}{6}} = \frac{\sqrt{6} + 2\sqrt{3}}{2\sqrt{3} - \sqrt{6}}$$ 69. Rationalize denominator: $$= \frac{\sqrt{6} + 2\sqrt{3}}{2\sqrt{3} - \sqrt{6}} \cdot \frac{2\sqrt{3} + \sqrt{6}}{2\sqrt{3} + \sqrt{6}} = \frac{(\sqrt{6} + 2\sqrt{3})(2\sqrt{3} + \sqrt{6})}{(2\sqrt{3})^2 - (\sqrt{6})^2}$$ 70. Calculate denominator: $$= 4 \cdot 3 - 6 = 12 - 6 = 6$$ 71. Numerator: Expand: $$= \sqrt{6} \cdot 2 \sqrt{3} + \sqrt{6} \cdot \sqrt{6} + 2\sqrt{3} \cdot 2 \sqrt{3} + 2\sqrt{3} \cdot \sqrt{6} = 2 \sqrt{18} + 6 + 4 \cdot 3 + 2 \sqrt{18} = 2 \cdot 3 \sqrt{2} + 6 + 12 + 2 \cdot 3 \sqrt{2} = 6 \sqrt{2} + 18 + 6 \sqrt{2} = 12 \sqrt{2} + 18$$ 72. Final: $$\tan(A-B) = \frac{12 \sqrt{2} + 18}{6} = 3 \sqrt{2} + 3$$ 73. Determine quadrant of $A-B$: since numerator in sine is positive, denominator in cosine is positive (approx), check signs: $\sin(A-B) > 0$, $\cos(A-B) > 0$ so $A-B$ is in quadrant I. --- 74. The problem: Given $\sin \alpha = \frac{4}{5}$ and $\cos \beta = \frac{5}{13}$, find $\sin(\alpha + \beta) + \sin(\alpha - \beta)$. 75. Use sum-to-product formula: $$\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2 \sin \alpha \cos \beta$$ 76. Substitute: $$= 2 \cdot \frac{4}{5} \cdot \frac{5}{13} = \frac{40}{65} = \frac{8}{13}$$ --- 77. The problem: Given $\sin \alpha = \frac{2}{3}$, $\alpha$ in quadrant II and $\cos \beta = \frac{3}{4}$, find $\cos(\alpha+\beta) + \cos(\alpha - \beta)$. 78. Use sum-to-product formula: $$\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2 \cos \alpha \cos \beta$$ 79. Since $\sin \alpha = 2/3$ and $\alpha$ in QII, $\cos \alpha$ is negative: $$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \frac{4}{9}} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$$ 80. Substitute values: $$= 2 \cdot \left(-\frac{\sqrt{5}}{3}\right) \cdot \frac{3}{4} = 2 \cdot \left(-\frac{\sqrt{5}}{3} \cdot \frac{3}{4}\right) = 2 \cdot \left(-\frac{\sqrt{5}}{4}\right) = -\frac{\sqrt{5}}{2}$$ --- 81. The problem: If $A$ and $B$ are acute angles (degrees) such that $\csc A = \sqrt{17}$ and $\csc B = \sqrt{\frac{34}{3}}$, find $A + B$. 82. From $\csc A = \sqrt{17}$, $\sin A = \frac{1}{\sqrt{17}}$. 83. Find $A$: $$A = \sin^{-1}\left(\frac{1}{\sqrt{17}}\right)$$ 84. From $\csc B = \sqrt{\frac{34}{3}}$, $$\sin B = \frac{1}{\sqrt{34/3}} = \sqrt{\frac{3}{34}}$$ 85. Calculate $B$: $$B = \sin^{-1} \left(\sqrt{\frac{3}{34}}\right)$$ 86. Sum: $$A + B = \sin^{-1}\left(\frac{1}{\sqrt{17}}\right) + \sin^{-1} \left(\sqrt{\frac{3}{34}}\right)$$ 87. Exact values are not standard angles. Numerically, $A \approx 0.242$ rad, $B \approx 0.305$ rad, so sum $\approx 0.547$ rad $\approx 31.35^\circ$. --- 88. The problem: If $\tan(x+y) = \frac{1}{3}$ and $\tan y = \frac{1}{2}$, find $\tan x$. 89. Use formula for tangent of sum: $$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$ 90. Substitute known values: $$\frac{1}{3} = \frac{\tan x + \frac{1}{2}}{1 - \tan x \cdot \frac{1}{2}}$$ 91. Multiply both sides by denominator: $$\frac{1}{3} \left(1 - \frac{\tan x}{2}\right) = \tan x + \frac{1}{2}$$ 92. Multiply out left: $$\frac{1}{3} - \frac{\tan x}{6} = \tan x + \frac{1}{2}$$ 93. Group $\tan x$ terms: $$-\frac{\tan x}{6} - \tan x = \frac{1}{2} - \frac{1}{3}$$ 94. Left side: $$- \frac{\tan x}{6} - \tan x = - \frac{\tan x}{6} - \frac{6 \tan x}{6} = -\frac{7 \tan x}{6}$$ 95. Right side: $$\frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$$ 96. So, $$-\frac{7 \tan x}{6} = \frac{1}{6} \implies -7 \tan x = 1 \implies \tan x = -\frac{1}{7}$$ --- 97. The problem: Evaluate $$\frac{\tan \frac{\pi}{9} + \tan \frac{23\pi}{36}}{1 - \tan \frac{\pi}{9} \tan \frac{23\pi}{36}}$$ 98. Recognize this as tangent addition formula: $$\tan(a) + \tan(b) \over 1 - \tan a \tan b = \tan(a+b)$$ 99. Sum angles: $$\frac{\pi}{9} + \frac{23 \pi}{36} = \frac{4\pi}{36} + \frac{23 \pi}{36} = \frac{27 \pi}{36} = \frac{3 \pi}{4}$$ 100. So, $$\frac{\tan \frac{\pi}{9} + \tan \frac{23\pi}{36}}{1 - \tan \frac{\pi}{9} \tan \frac{23\pi}{36}} = \tan \frac{3 \pi}{4} = -1$$ --- 101. The problem: Establish the identity $$\sin(A+B+C) = \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C$$ 102. Use addition formula twice: $$\sin(A+B+C) = \sin((A+B) + C) = \sin(A+B) \cos C + \cos(A+B) \sin C$$ 103. Expand $\sin(A+B)$: $$\sin A \cos B + \cos A \sin B$$ 104. Expand $\cos(A+B)$: $$\cos A \cos B - \sin A \sin B$$ 105. Substitute: $$= (\sin A \cos B + \cos A \sin B) \cos C + (\cos A \cos B - \sin A \sin B) \sin C$$ 106. Distribute: $$= \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C$$ 107. Identity established. --- 108. The problem: Prove $$\sin 2\theta = 2 \sin \theta \cos \theta$$ 109. Use double-angle formula: $$\sin 2\theta = \sin(\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta = 2 \sin \theta \cos \theta$$ 110. Identity proven. --- 111. The problem: Prove $$\cot 2\theta = \frac{\cot^2 \theta - 1}{2 \cot \theta}$$ 112. Start from definition: $$\cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}$$ 113. Use identities: $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$ $$\sin 2\theta = 2 \sin \theta \cos \theta$$ 114. Substitute: $$\cot 2\theta = \frac{\cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta}$$ 115. Write numerator as: $$\cos^2 \theta - \sin^2 \theta = (\frac{\cos \theta}{\sin \theta})^2 \sin^2 \theta - \sin^2 \theta = \sin^2 \theta (\cot^2 \theta -1)$$ but better to factor: 116. Divide numerator and denominator by $\sin^2 \theta$: $$= \frac{\frac{\cos^2 \theta}{\sin^2 \theta} -1}{2 \frac{\sin \theta \cos \theta}{\sin^2 \theta}} = \frac{\cot^2 \theta -1}{2 \cot \theta}$$ 117. Identity proven.