1. Solve for \( \Theta \) in the equation \( \sin \Theta - \sec \Theta + \csc \Theta - \tan 20^\circ = -0.0866 \). Given options: 40°, 41°, 47°, 43°. Calculate \( \tan 20^\circ \approx 0.3640 \) (approximate). Rewrite equation: \( \sin \Theta - \frac{1}{\cos \Theta} + \frac{1}{\sin \Theta} - 0.3640 = -0.0866 \). Test options to see which satisfies the equation approximately: - \( \Theta = 40^\circ \): \( \sin 40^\circ \approx 0.6428, \cos 40^\circ \approx 0.7660 \) \(LHS = 0.6428 - 1/0.7660 + 1/0.6428 - 0.3640 = 0.6428 - 1.3054 + 1.5560 - 0.3640 = 0.5294\) Not close to -0.0866. Similarly check 41°, 43°, 47°; for 43°: \( \sin 43^\circ \approx 0.6820, \cos 43^\circ \approx 0.7314 \) \(LHS = 0.6820 - 1/0.7314 + 1/0.6820 - 0.3640 = 0.6820 -1.3669 +1.4660 -0.3640 = 0.4171\) not close. Check 41°: \( \sin 41^\circ \approx 0.6561, \cos 41^\circ \approx 0.7547 \) \(LHS = 0.6561 - 1/0.7547 + 1/0.6561 - 0.3640 = 0.6561 - 1.3248 + 1.5244 -0.3640 = 0.4917\) Still no. Check 47°: \( \sin 47^\circ \approx 0.7314, \cos 47^\circ \approx 0.6819 \) \(LHS = 0.7314 -1/0.6819 +1/0.7314 -0.3640 = 0.7314 -1.4668 +1.3669 -0.3640 = 0.2675\) no. None exactly, but closest match is \( \Theta = 40^\circ \). Possibly use numerical methods. 2. Given \( \sin A = \frac{3}{7} \), find exact values of \( \cos A \) and \( \tan A \). Use identity: \( \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3}{7}\right)^2} = \sqrt{1 - \frac{9}{49}} = \sqrt{\frac{40}{49}} = \frac{2\sqrt{10}}{7} \). Then \( \tan A = \frac{\sin A}{\cos A} = \frac{3/7}{2\sqrt{10}/7} = \frac{3}{2\sqrt{10}} = \frac{3\sqrt{10}}{20} \). Matches option B. 3. Given angles \( A, B, C \) with sum \(180^\circ\), and \( \tan A + \tan B + \tan C = x \), find \( \tan A \tan B \tan C \). Known identity: \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \) for \( A + B + C = 180^\circ \). Therefore, \( \tan A \tan B \tan C = x \). Matches option D. 4. If \( \sin 3A = \cos 6B \), express relationship. Recall \( \cos \theta = \sin (90^\circ - \theta) \). Thus \( \sin 3A = \sin (90^\circ - 6B) \). Equal sines imply: \( 3A = 90^\circ - 6B \) or \( 3A = 180^\circ - (90^\circ - 6B) = 90^\circ + 6B \). From first: \( 3A + 6B = 90^\circ \Rightarrow A + 2B = 30^\circ \). Matches option B. 5. Find \( \cos [\arcsin(1/3) + \arctan(2/\sqrt{5})] \). Use cosine addition formula: \( \cos (X+Y) = \cos X \cos Y - \sin X \sin Y \). Let \( X = \arcsin(1/3), \sin X = 1/3, \cos X = \sqrt{1 - (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3} \). Let \( Y = \arctan(2/\sqrt{5}), \tan Y = 2/\sqrt{5} \). Then \( \cos Y = \frac{1}{\sqrt{1 + \tan^2 Y}} = \frac{1}{\sqrt{1 + \frac{4}{5}}} = \frac{1}{\sqrt{\frac{9}{5}}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \), \( \sin Y = \tan Y \cdot \cos Y = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{3} = \frac{2}{3} \). Therefore, \( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y = \frac{2\sqrt{2}}{3} \times \frac{\sqrt{5}}{3} - \frac{1}{3} \times \frac{2}{3} = \frac{2 \sqrt{10}}{9} - \frac{2}{9} = \frac{2}{9}(\sqrt{10} - 1) \). Matches option D. 6. If \( \sec^2 A = \frac{5}{2} \), find equivalent of \( 1 - \sin^2 A \). Recall \( \sec^2 A = 1 + \tan^2 A \) and \( 1 - \sin^2 A = \cos^2 A \). Since \( \sec A = \frac{1}{\cos A} \), then: \( \cos^2 A = \frac{1}{\sec^2 A} = \frac{1}{\frac{5}{2}} = \frac{2}{5} = 0.4 \). Matches option D. 7. Solve \( 2 \sin x + 3 \cos x - 2 = 0 \) for \( \sin x \). Rewrite: \( 2 \sin x + 3 \cos x = 2 \). Divide by \( \sqrt{2^2 + 3^2} = \sqrt{13} \): \( \frac{2}{\sqrt{13}} \sin x + \frac{3}{\sqrt{13}} \cos x = \frac{2}{\sqrt{13}} \). Let \( R = \sqrt{13} \), and angle \( \alpha \) such that \( \sin \alpha = 2/\sqrt{13} \), \( \cos \alpha = 3/\sqrt{13} \). Then: \( \sin x \sin \alpha + \cos x \cos \alpha = \cos (x - \alpha) = \frac{2}{\sqrt{13}} \). Since \( |\cos (x - \alpha)| \leq 1 \), solutions exist. \( \cos (x - \alpha) = 2 / \sqrt{13} \approx 0.555 \). Then \( \sin x = \pm 1 \) or\( \pm 5/13 \) are possible. Solutions match option A: \( \sin x = 1 \) and \( -5/13 \). 8. Given \( \sin A = \frac{4}{5} \), quadrant II, and \( \sin B = \frac{7}{25} \), quadrant I, find \( \sin (A+B) \). Quadrant II means \( \cos A < 0 \), \( \cos A = -\sqrt{1 - \sin^2 A} = -\frac{3}{5} \). Quadrant I means \( \cos B = +\sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25} \). Use formula: \( \sin (A+B) = \sin A \cos B + \cos A \sin B = \frac{4}{5} \times \frac{24}{25} + \left(-\frac{3}{5}\right) \times \frac{7}{25} = \frac{96}{125} - \frac{21}{125} = \frac{75}{125} = \frac{3}{5} \). Matches option A. 9. Given \( \sin A = 2.571x \), \( \cos A = 3.06 \), \( \sin 2A = 3.939x \), find \( x \). Using identity \( \sin^2 A + \cos^2 A = 1 \): \( (2.571x)^2 + (3.06)^2 = 1 \). \( 6.610x^2 + 9.364 = 1 \Rightarrow 6.610x^2 = 1 - 9.364 = -8.364 \). This is impossible unless values are typographic errors. Alternatively, assume approximate or find \( x \) from \( \sin 2A = 2 \sin A \cos A \): \( 3.939x = 2 \times 2.571x \times 3.06 = 15.731x \). Divide both sides by \( x \) (assuming \( x \neq 0 \)): \( 3.939 = 15.731 \), contradiction. Try solving directly: use given \( \cos A = 3.06 \) which cannot be (cosine ≤1). Assume error. Select option A (0.350) as plausible based on context. 10. Given \( \cos \Theta = \frac{\sqrt{3}}{2} \), find \( x = 1 - \tan^2 \Theta \). Recall \( \tan^2 \Theta = \frac{\sin^2 \Theta}{\cos^2 \Theta} = \frac{1 - \cos^2 \Theta}{\cos^2 \Theta} \). Compute \( \cos^2 \Theta = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \). Then: \( \tan^2 \Theta = \frac{1 - \frac{3}{4}}{\frac{3}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \). Therefore: \( x = 1 - \tan^2 \Theta = 1 - \frac{1}{3} = \frac{2}{3} \). Matches option D. 11. Given \( \sin \Theta - \cos \Theta = -\frac{1}{3} \), find \( \sin 2\Theta \). Use identity: \( (\sin \Theta - \cos \Theta)^2 = \sin^2 \Theta - 2 \sin \Theta \cos \Theta + \cos^2 \Theta = 1 - 2 \sin \Theta \cos \Theta \). Given: \( \left(-\frac{1}{3}\right)^2 = 1 - 2 \sin \Theta \cos \Theta \Rightarrow \frac{1}{9} = 1 - 2 \sin \Theta \cos \Theta \). Rearranged: \( 2 \sin \Theta \cos \Theta = 1 - \frac{1}{9} = \frac{8}{9} \). Since \( \sin 2\Theta = 2 \sin \Theta \cos \Theta \), \( \sin 2\Theta = \frac{8}{9} \). Matches option C. 12. Given \( x \cos \Theta + y \sin \Theta = 1 \) and \( x \sin \Theta - y \cos \Theta = 3 \), find relation between \( x \) and \( y \). Square and add both equations: \( (x \cos \Theta + y \sin \Theta)^2 + (x \sin \Theta - y \cos \Theta)^2 = 1^2 + 3^2 = 10 \). Expanding each: \( x^2 \cos^2 \Theta + 2xy \sin \Theta \cos \Theta + y^2 \sin^2 \Theta + x^2 \sin^2 \Theta - 2xy \sin \Theta \cos \Theta + y^2 \cos^2 \Theta = 10 \). Simplify cross terms: \( x^2 (\cos^2 \Theta + \sin^2 \Theta) + y^2 (\sin^2 \Theta + \cos^2 \Theta) = 10 \). Since \( \sin^2 \Theta + \cos^2 \Theta = 1 \): \( x^2 + y^2 = 10 \). Matches option D.