Q41 / 75 · JEE Main 2026 (04 April Shift 1) · Mathematics

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

  • A
    $3$
  • B
    $-1$
  • Correct answer
    C
    $1$
  • D
    $2$

Solution

Let $y = y_1 + y_2$

$y_1 = \tan^{-1}\left(\dfrac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right)$

Dividing the numerator and denominator by $4\cos x$:

$y_1 = \tan^{-1}\left(\dfrac{\dfrac{3}{4} - \tan x}{1 + \dfrac{3}{4}\tan x}\right)$

Using the formula $\tan^{-1}\left(\dfrac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B$:

$y_1 = \tan^{-1}\left(\dfrac{3}{4}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\dfrac{3}{4}\right) - x$

Differentiating with respect to $x$:

$\dfrac{dy_1}{dx} = -1$

Now, $y_2 = 2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1-x^2}}\right)$

Substituting $x = \sin \theta$:

$y_2 = 2\tan^{-1}\left(\dfrac{\sin \theta}{1+\cos \theta}\right)$

$y_2 = 2\tan^{-1}\left(\dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}\right)$

$y_2 = 2\tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = 2\left(\dfrac{\theta}{2}\right) = \theta = \sin^{-1}x$

Differentiating with respect to $x$:

$\dfrac{dy_2}{dx} = \dfrac{1}{\sqrt{1-x^2}}$

At $x = \dfrac{\sqrt{3}}{2}$:

$\dfrac{dy_2}{dx} = \dfrac{1}{\sqrt{1 - \left(\dfrac{\sqrt{3}}{2}\right)^2}} = \dfrac{1}{\sqrt{1 - \dfrac{3}{4}}} = \dfrac{1}{\sqrt{\dfrac{1}{4}}} = 2$

Therefore, $\dfrac{dy}{dx} = \dfrac{dy_1}{dx} + \dfrac{dy_2}{dx} = -1 + 2 = 1$

Answer: $1$

Hence, the correct answer is option: 'C'.

Relevant Questions