Question

The derivative of \[ \sin^{-1} \left(\frac{2x}{1 + x^2} \right) \] with respect to \[ \cos^{-1} \left(\frac{1 - x^2}{1 + x^2} \right) \] is equal to:

• A. \( 1 \)
• B. \( -1 \)
• C. \( 2 \)
• D. None of these

Hint:

For inverse trigonometric functions, recognize standard identities like \( \sin^{-1} \left(\frac{2x}{1 + x^2} \right) = 2 \tan^{-1} x \).

Answer 1

The Correct Option is A

Step 1: Setting variables.
Let \[ y = \sin^{-1} \left(\frac{2x}{1 + x^2} \right), \quad z = \cos^{-1} \left(\frac{1 - x^2}{1 + x^2} \right) \] Step 2: Recognizing trigonometric identities.
We use the well-known identity: \[ \sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} \] which implies \[ \sin^{-1} \left(\frac{2x}{1 + x^2} \right) = 2 \tan^{-1} x \] Similarly, we use \[ \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \] which gives \[ \cos^{-1} \left(\frac{1 - x^2}{1 + x^2} \right) = 2 \tan^{-1} x \] Step 3: Differentiating both functions.
\[ \frac{dy}{dx} = \frac{d}{dx} (2 \tan^{-1} x) = \frac{2}{1 + x^2} \] \[ \frac{dz}{dx} = \frac{d}{dx} (2 \tan^{-1} x) = \frac{2}{1 + x^2} \] Step 4: Computing \( \frac{dy}{dz} \).
\[ \frac{dy}{dz} = \frac{\frac{2}{1 + x^2}}{\frac{2}{1 + x^2}} = 1 \] Thus, the correct answer is (A) \( 1 \).