Question

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

• A. $\dfrac{1}{2}$
• B. $2$
• C. $-2$
• D. $-\dfrac{1}{2}$
• E. $1$

Hint:

When simplifying trigonometric expressions, always convert everything to $\sin$ and $\cos$ first. Use $1 + \cos 2\theta = 2\cos^2\theta$ and $\sin^2\theta - 1 = -\cos^2\theta$ to cancel terms cleanly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We simplify $x$ and $y$ using standard trigonometric identities, then compute the ratio $y/x$.

Step 2: Detailed Explanation:
Recall $1 + \cos 2\theta = 2\cos^2\theta$, $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$.
\[ x = \frac{2\cos^2\theta}{\dfrac{\sin\theta - 1}{\cos\theta}} = \frac{2\cos^3\theta}{\sin\theta - 1} \] \[ y = \frac{\dfrac{\sin\theta+1}{\cos\theta}}{\dfrac{1}{\cos^2\theta}} = \frac{(\sin\theta+1)\cos\theta}{1} = (\sin\theta+1)\cos\theta \] Now, \[ \frac{y}{x} = \frac{(\sin\theta+1)\cos\theta \cdot (\sin\theta-1)}{2\cos^3\theta} = \frac{(\sin^2\theta - 1)}{2\cos^2\theta} = \frac{-\cos^2\theta}{2\cos^2\theta} = -\frac{1}{2} \]

Step 3: Final Answer:
$\dfrac{y}{x} = -\dfrac{1}{2}$