Question

If \( x \cos(p + y) + \cos p \sin(p + y) = 0 \), prove that \( \cos p \frac{dy}{dx} = -\cos^2(p + y) \), where \( p \) is a constant.

Hint:

For trigonometric identities involving derivatives, simplify using tangent and secant relationships before differentiating.

Answer 1

Step 1: Rearrange the given equation
The given equation is: \[ x \cos(p + y) + \cos p \sin(p + y) = 0. \] Divide throughout by \( \cos(p + y) \): \[ x + \cos p \tan(p + y) = 0. \] This implies: \[ \tan(p + y) = -\frac{x}{\cos p}. \] 
Step 2: Differentiate with respect to \( x \)
Differentiate both sides with respect to \( x \): \[ \sec^2(p + y) \cdot \frac{d}{dx}(p + y) = -\frac{d}{dx}\left(\frac{x}{\cos p}\right). \] Simplify: \[ \sec^2(p + y) \frac{dy}{dx} = -\frac{1}{\cos p}. \] 
Step 3: Express \( \frac{dy}{dx} \) in terms of \( \cos^2(p + y) \)
Using \( \sec^2(p + y) = \frac{1}{\cos^2(p + y)} \), we get: \[ \frac{1}{\cos^2(p + y)} \cdot \frac{dy}{dx} = -\frac{1}{\cos p}. \] Multiply through by \( \cos^2(p + y) \): \[ \frac{dy}{dx} = -\cos^2(p + y) \cdot \cos p. \] 
Step 4: Conclude the result
\[ \cos p \frac{dy}{dx} = -\cos^2(p + y). \]