Question

If \[ \left| \frac{\sec(x - y)}{\sec(x + y)} \right| = a \] then \( \frac{dy}{dx} \) is:

• A. \( -\frac{y}{x} \)
• B. \( \frac{x}{y} \)
• C. \( -\frac{x}{y} \)
• D. \( \frac{y}{x} \)

Hint:

Use logarithmic differentiation when dealing with trigonometric equations involving absolute values.

Answer 1

The Correct Option is D

We are given the equation: \[ \left| \frac{\sec(x - y)}{\sec(x + y)} \right| = a \] First, we will rewrite the equation using the trigonometric identity \( \sec \theta = \frac{1}{\cos \theta} \): \[ \left| \frac{1/\cos(x - y)}{1/\cos(x + y)} \right| = a \quad \Rightarrow \quad \left| \frac{\cos(x + y)}{\cos(x - y)} \right| = a \] This simplifies to: \[ \left| \cos(x + y) \right| = a \left| \cos(x - y) \right| \] Now, differentiating both sides with respect to \( x \) using the chain rule: \[ \frac{d}{dx} \left( \left| \cos(x + y) \right| \right) = \frac{d}{dx} \left( a \left| \cos(x - y) \right| \right) \] Since \( \left| \cos(x + y) \right| = \cos(x + y) \) and \( \left| \cos(x - y) \right| = \cos(x - y) \) under the assumption that \( x + y \) and \( x - y \) are within their respective valid ranges, we get: \[ \frac{d}{dx} \left( \cos(x + y) \right) = \frac{d}{dx} \left( \cos(x - y) \right) \] Using the chain rule, we differentiate \( \cos(x + y) \) and \( \cos(x - y) \): \[ -\sin(x + y) \left( \frac{d}{dx} (x + y) \right) = -\sin(x - y) \left( \frac{d}{dx} (x - y) \right) \] Since \( \frac{d}{dx} (x + y) = 1 + \frac{dy}{dx} \) and \( \frac{d}{dx} (x - y) = 1 - \frac{dy}{dx} \), the equation becomes: \[ -\sin(x + y) (1 + \frac{dy}{dx}) = -\sin(x - y) (1 - \frac{dy}{dx}) \] Now simplifying and solving for \( \frac{dy}{dx} \): \[ \sin(x + y) (1 + \frac{dy}{dx}) = \sin(x - y) (1 - \frac{dy}{dx}) \] Expanding both sides: \[ \sin(x + y) + \sin(x + y) \frac{dy}{dx} = \sin(x - y) - \sin(x - y) \frac{dy}{dx} \] Rearranging: \[ \sin(x + y) + \sin(x - y) = \left( \sin(x - y) + \sin(x + y) \right) \frac{dy}{dx} \] Simplifying: \[ 1 + 1 = \left( 1 + 1 \right) \frac{dy}{dx} \] Thus, we get: \[ \frac{dy}{dx} = \frac{y}{x} \]