Question

If \( y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \), then \(\frac{dy}{dx}\) at \(x = \frac{\sqrt{5}}{2}\) is equal to:

• A. 3
• B. -1
• C. 1
• D. 2

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to simplify each term of the inverse trigonometric function expression using standard substitution and algebraic manipulations before differentiating.
Step 2: Key Formula or Approach:
For the first term: Divide numerator and denominator by \(\cos x\) to use the identity \(\tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B\).
For the second term: Substitute \(x = \sin \theta\) to simplify \(\frac{x}{1 + \sqrt{1 - x^2}}\) using half-angle formulas.
Step 3: Detailed Explanation:
Let \(y = u + v\).
**First term (\(u\)):**
\[ u = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) = \tan^{-1}\left(\frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4}\tan x}\right) \] Using the identity, this simplifies to:
\[ u = \tan^{-1}\left(\frac{3}{4}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\frac{3}{4}\right) - x \] **Second term (\(v\)):**
\[ v = 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \] Let \(x = \sin \theta \implies \theta = \sin^{-1}x\).
\[ \frac{x}{1 + \sqrt{1 - x^2}} = \frac{\sin \theta}{1 + \cos \theta} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2) \] Thus:
\[ v = 2\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = 2\left(\frac{\theta}{2}\right) = \theta = \sin^{-1}x \] So, the full simplified function is:
\[ y = \tan^{-1}\left(\frac{3}{4}\right) - x + \sin^{-1}x \] Now, differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 0 - 1 + \frac{1}{\sqrt{1 - x^2}} \] Evaluate at \(x = \frac{\sqrt{3}}{2}\):
\[ \left. \frac{dy}{dx} \right|_{x=\frac{\sqrt{3}}{2}} = -1 + \frac{1}{\sqrt{1 - (\frac{\sqrt{3}}{2})^2}} = -1 + \frac{1}{\sqrt{1 - \frac{3}{4}}} \] \[ = -1 + \frac{1}{\sqrt{\frac{1}{4}}} = -1 + \frac{1}{\frac{1}{2}} = -1 + 2 = 1 \] Step 4: Final Answer:
The value of the derivative is 1.