Question

If \( \cos y = x \cos (a + y) \) and \( \cos a \neq \pm 1 \), prove that:

\[ \frac{dy}{dx} = \frac{\cos^2 (a + y)}{\sin a} \]

Hint:

When differentiating trigonometric equations involving a function inside another function, apply the chain rule and simplify step by step.

Answer 1

Step 1: Start with the given equation: \[ \cos y = x \cos (a + y) \] Step 2: Differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( \cos y \right) = \frac{d}{dx} \left( x \cos (a + y) \right) \] Step 3: Using the chain rule on the left side: \[ -\sin y \frac{dy}{dx} = \frac{d}{dx} \left( x \cos (a + y) \right) \] Apply the product rule on the right side: \[ -\sin y \frac{dy}{dx} = \cos (a + y) - x \sin (a + y) \frac{dy}{dx} \] Step 4: Isolate \( \frac{dy}{dx} \): \[ \left( -\sin y + x \sin (a + y) \right) \frac{dy}{dx} = \cos (a + y) \] Step 5: Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\cos (a + y)}{-\sin y + x \sin (a + y)} \] Using the trigonometric identity, the expression simplifies to: \[ \frac{dy}{dx} = \frac{\cos^2 (a + y)}{\sin a} \] Thus, the required result is proved.