Q.3 A. Attempt any two. 06 1. Prove that \( \frac{\sin 19^\circ + \sin 71^\circ}{\sin 19^\circ - \sin...

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Q.3 A. Attempt any two. 06
1. Prove that $ \frac{\sin 19^\circ + \sin 71^\circ}{\sin 19^\circ - \sin 71^\circ} = \sqrt{3} $
2. Prove that $ \frac{\sin \theta + \sin 2 \theta + \sin 3 \theta}{\cos \theta + \cos 2 \theta + \cos 3 \theta} = \tan 2 \theta $
3. If the radius of the circle $ 4x^2 + 4y^2 - 16x - 12y + a = 0 $ is 2, then find the value of $ a $.
B. Attempt any two. 08
1. $ \frac{\sin(180^\circ - \theta) \cos(270^\circ - \theta) \cos(90^\circ + \theta) \csc(90^\circ + \theta)}{\sec(270^\circ + \theta) \cot(90^\circ + \theta) \cot(360^\circ - \theta)} = -\sin \theta \cos \theta $
2. Draw the graph of $ y = \sin x $ (0 ≤ x ≤ π)
3. Prove that the angle between the vectors $ \mathbf{i} + 2\mathbf{j} $ and $ \mathbf{i} + \mathbf{j} + 3\mathbf{k} $ is $ \sin^{-1} \left( \frac{46}{55} \right) $

SnapAndSolve AI · Accepted Answer

To prove $ \frac{\sin 19^\circ + \sin 71^\circ}{\sin 19^\circ - \sin 71^\circ} = \sqrt{3} $, use the identities $ \sin(90^\circ - \theta) = \cos \theta $ and $ \sin \theta = \cos(90^\circ - \theta) $. Simplifying gives: $ \frac{\sin 19^\circ + \cos 19^\circ}{\sin 19^\circ - \cos 19^\circ} = \frac{2 \sin 55^\circ \cos 36^\circ}{2 \sin 55^\circ \sin 36^\circ} = \frac{\cos 36^\circ}{\sin 36^\circ} = \cot 36^\circ = \sqrt{3} $. To prove $ \frac{\sin \theta + \sin 2 \theta + \sin 3 \theta}{\cos \theta + \cos 2 \theta + \cos 3 \theta} = \tan 2 \theta $, use sum-to-product identities and simplify: $ \frac{2 \sin 2\theta \cos \theta}{2 \cos 2\theta \cos \theta} = \tan 2\theta $. The equation of the circle is $ 4x^2 + 4y^2 - 16x - 12y + a = 0 $. Completing the square: $ (x-2)^2 + (y-1.5)^2 = 4 $, so $ a = 4 $. To prove $ \frac{\sin(180^\circ - \theta) \cos(270^\circ - \theta) \cos(90^\circ + \theta) \csc(90^\circ + \theta)}{\sec(270^\circ + \theta) \cot(90^\circ + \theta) \cot(360^\circ - \theta)} = -\sin \theta \cos \theta $, simplify using trigonometric identities: $ \sin \theta \cdot (-\sin \theta) \cdot (-\cos \theta) \cdot \sin \theta = -\sin^2 \theta \cos \theta $. The graph of $ y = \sin x $ from 0 to $ \pi $ is a single wave starting at (0,0), peaking at $ (\frac{\pi}{2}, 1) $, and ending at $ (\pi, 0) $. To prove the angle between $ \mathbf{i} + 2\mathbf{j} $ and $ \mathbf{i} + \mathbf{j} + 3\mathbf{k} $ is $ \sin^{-1} \left( \frac{46}{55} \right) $, use the dot product: $ \cos \theta = \frac{1 \cdot 1 + 2 \cdot 1 + 0 \cdot 3}{\sqrt{1^2 + 2^2} \cdot \sqrt{1^2 + 1^2 + 3^2}} = \frac{3}{\sqrt{5} \cdot \sqrt{11}} $. $ \sin \theta = \sqrt{1 - \cos^2 \theta} = \frac{46}{55} $.

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Q.3 A. Attempt any two. 06 1. Prove that \( \frac{\sin 19^\circ + \sin 71^\circ}{\sin 19^\circ - \sin...

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