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JEE Main 20264 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 4 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

If y = ^ -1 ((3 x - 4 x)/(4 x + 3 x)) + 2 ^ -1 ( x 1+ 1-x^2 ) ,...

If y=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)+2tan⁡−1(x1+1−x2)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 dydx\dfrac{dy}{dx} at x=32x = \dfrac{\sqrt{3}}{2} is equal to:

Options

  1. A33
  2. B−1-1
  3. C11
  4. D22

Correct answer

C. 11

Step-by-step solution

Let y=y1+y2y = y_1 + y_2

y1=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)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 4cos⁡x4\cos x:

y1=tan⁡−1(34−tan⁡x1+34tan⁡x)y_1 = \tan^{-1}\left(\dfrac{\dfrac{3}{4} - \tan x}{1 + \dfrac{3}{4}\tan x}\right)

Using the formula tan⁡−1(A−B1+AB)=tan⁡−1A−tan⁡−1B\tan^{-1}\left(\dfrac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B:

y1=tan⁡−1(34)−tan⁡−1(tan⁡x)=tan⁡−1(34)−xy_1 = \tan^{-1}\left(\dfrac{3}{4}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\dfrac{3}{4}\right) - x

Differentiating with respect to xx:

dy1dx=−1\dfrac{dy_1}{dx} = -1

Now, y2=2tan⁡−1(x1+1−x2)y_2 = 2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1-x^2}}\right)

Substituting x=sin⁡θx = \sin \theta:

y2=2tan⁡−1(sin⁡θ1+cos⁡θ)y_2 = 2\tan^{-1}\left(\dfrac{\sin \theta}{1+\cos \theta}\right)

y2=2tan⁡−1(2sin⁡(θ/2)cos⁡(θ/2)2cos⁡2(θ/2))y_2 = 2\tan^{-1}\left(\dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}\right)

y2=2tan⁡−1(tan⁡θ2)=2(θ2)=θ=sin⁡−1xy_2 = 2\tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = 2\left(\dfrac{\theta}{2}\right) = \theta = \sin^{-1}x

Differentiating with respect to xx:

dy2dx=11−x2\dfrac{dy_2}{dx} = \dfrac{1}{\sqrt{1-x^2}}

At x=32x = \dfrac{\sqrt{3}}{2}:

dy2dx=11−(32)2=11−34=114=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, dydx=dy1dx+dy2dx=−1+2=1\dfrac{dy}{dx} = \dfrac{dy_1}{dx} + \dfrac{dy_2}{dx} = -1 + 2 = 1

Answer: 11
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