Question

\(x = \frac{1 + \cos 2\theta}{\tan \theta - \sec \theta}\) and \(y = \frac{\tan \theta + \sec \theta}{\sec^2 \theta}\), then \(\frac{y}{x} =\)

• A. \(\frac{1}{2}\)
• B. 2
• C. -2
• D. \(-\frac{1}{2}\)
• E. 1

Hint:

\(\sin^2 \theta - 1 = -\cos^2 \theta\).

Answer 1

The Correct Option is D

Step 1: Concept:
• Use standard trigonometric identities: \[ 1 + \cos 2\theta = 2\cos^2 \theta \] \[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \sec \theta = \frac{1}{\cos \theta} \]

Step 2: Simplify \(x\) and \(y\):
• Simplifying \(x\): \[ x = \frac{2\cos^2 \theta}{\frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta}} = \frac{2\cos^2 \theta}{\frac{\sin \theta - 1}{\cos \theta}} = \frac{2\cos^3 \theta}{\sin \theta - 1} \]
• Simplifying \(y\): \[ y = \frac{\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}}{\frac{1}{\cos^2 \theta}} = \frac{\frac{\sin \theta + 1}{\cos \theta}}{\frac{1}{\cos^2 \theta}} = (\sin \theta + 1)\cos \theta \]

Step 3: Find \(\frac{y}{x}\):
• Compute ratio: \[ \frac{y}{x} = \frac{(\sin \theta + 1)\cos \theta}{\frac{2\cos^3 \theta}{\sin \theta - 1}} \]
• Simplify: \[ = \frac{(\sin \theta + 1)(\sin \theta - 1)}{2\cos^2 \theta} = \frac{\sin^2 \theta - 1}{2\cos^2 \theta} \]
• Using identity \(\sin^2 \theta - 1 = -\cos^2 \theta\): \[ \frac{y}{x} = \frac{-\cos^2 \theta}{2\cos^2 \theta} = -\frac{1}{2} \]

Step 4: Final Answer:
• \[ \frac{y}{x} = -\frac{1}{2} \]