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Question
JEE Main
Mathematics

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 $x = \frac{\sqrt{3}}{2}$ is equal to:

The correct answer is
1

Derivative Calculation for Inverse Trigonometric Function

The problem requires finding the derivative $\frac{dy}{dx}$ of the function $y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$ and evaluating it at $x = \frac{\sqrt{3}}{2}$. We will differentiate each term separately.

Term 1 Differentiation

Consider the first term: $y_1 = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right)$.

  • Divide the numerator and denominator inside the $\tan^{-1}$ by $4\cos x$: $ \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} = \frac{3/4 - \tan x}{1 + (3/4)\tan x} $
  • Recognize the form $\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan(A - B)$, where $\tan A = 3/4$.
  • So, the expression inside $\tan^{-1}$ is $\tan(A - x)$, where $A = \tan^{-1}(3/4)$.
  • Therefore, $y_1 = \tan^{-1}(\tan(A - x))$. Assuming the principal value range, $y_1 = A - x$.
  • The derivative is $\frac{dy_1}{dx} = \frac{d}{dx}(A - x) = 0 - 1 = -1$.

Term 2 Differentiation

Consider the second term: $y_2 = 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$.

  • Let $x = \sin \theta$. Then $\theta = \sin^{-1}(x)$. This substitution is valid for $-1 \le x \le 1$.
  • Substitute $x = \sin \theta$ into the expression: $ \frac{\sin \theta}{1 + \sqrt{1 - \sin^2 \theta}} = \frac{\sin \theta}{1 + |\cos \theta|} $
  • Assuming $\cos \theta \ge 0$ (which corresponds to $-\pi/2 \le \theta \le \pi/2$, or $-1 \le x \le 1$), this becomes $\frac{\sin \theta}{1 + \cos \theta}$.
  • Using half-angle identities: $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1 + \cos \theta = 2\cos^2(\theta/2)$.
  • The expression simplifies to: $ \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2) $
  • So, $y_2 = 2\tan^{-1}(\tan(\theta/2))$.
  • Assuming the principal value range for $\tan^{-1}$, $\tan^{-1}(\tan(\theta/2)) = \theta/2$.
  • Therefore, $y_2 = 2(\theta/2) = \theta = \sin^{-1}(x)$.
  • The derivative is $\frac{dy_2}{dx} = \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}$.

Total Derivative and Evaluation

The total derivative is $\frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} = -1 + \frac{1}{\sqrt{1 - x^2}}$.

Now, evaluate the derivative at $x = \frac{\sqrt{3}}{2}$:

  • Calculate $x^2$: $x^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.
  • Calculate $1 - x^2$: $1 - x^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
  • Calculate $\sqrt{1 - x^2}$: $\sqrt{1 - x^2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
  • Calculate $\frac{1}{\sqrt{1 - x^2}}$: $\frac{1}{1/2} = 2$.
  • Substitute into the derivative expression: $ \frac{dy}{dx} \Big|_{x=\frac{\sqrt{3}}{2}} = -1 + 2 = 1 $

The value of the derivative at the given point is 1.

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