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:

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

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need \(\frac{dy}{dx}\) for the given function.

Step 2: Key Formula or Approach:
Simplify the argument using substitution \(a = r\cos\theta, b = r\sin\theta\).

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: \[ \frac{a\cos x - b\sin x}{b\cos x + a\sin x} = \cot(x+\theta) = \tan\left(\frac{\pi}{2} - x - \theta\right). \] Hence: \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - x - \theta\right)\right) = \frac{\pi}{2} - x - \theta. \] Differentiating: \[ \frac{dy}{dx} = -1. \]

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