Question

If \( y = \tan^{-1} \left\{ \frac{a \cos x - b \sin x}{b \cos x + a \sin x} \right\} \), then \( \frac{dy}{dx} = \)

• A. \( \frac{1}{1 + x^2} \)
• B. \( \frac{1}{\sqrt{1 - x^2}} \)
• C. \( -1 \)
• D. None of these

Hint:

For \(y = \tan^{-1}\left(\frac{a \cos x - b \sin x}{b \cos x + a \sin x}\right)\), the derivative is \(-1\) regardless of constants \(a\) and \(b\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need the derivative of the given inverse tangent function.

Step 2: Key Formula or Approach:
Let \(a = r \cos \theta\) and \(b = r \sin \theta\). Then simplify the argument.

Step 3: Detailed Explanation:
Let \(a = r \cos \theta\), \(b = r \sin \theta\). Then: \[ a \cos x - b \sin x = r \cos(x + \theta), \quad b \cos x + a \sin x = r \sin(x + \theta). \] Thus: \[ \frac{a \cos x - b \sin x}{b \cos x + a \sin x} = \cot(x + \theta) = \tan\left(\frac{\pi}{2} - x - \theta\right). \] Therefore: \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - x - \theta\right)\right) = \frac{\pi}{2} - x - \theta. \] Differentiating: \[ \frac{dy}{dx} = -1. \]

Step 4: Final Answer:
Option (C) is correct.