Start free

Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 4 April, Morning Shift — Question 21

Considering the principal values of the inverse trigonometric functions, sin⁡−1(32x+121−x2)\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^{2}}\right), −12<12-\frac{1}{2} < \frac{1}{\sqrt2}

  1. Option A:

    π4+sin⁡−1x\frac{\pi}{4}+\sin ^{-1} x

  2. Option B:

    π6+sin⁡−1x\frac{\pi}{6}+\sin ^{-1} x

    Correct
  3. Option C:

    5π6−sin⁡−1x\frac{5 \pi}{6}-\sin ^{-1} x

  4. Option D:

    −5π6−sin⁡−1x\frac{-5 \pi}{6}-\sin ^{-1} x

Answer: B

Step-by-step solution

Let θ=sin⁡−1x  ⟹  x=sin⁡θ, 1−x2=cos⁡θ\text{Let } \theta = \sin^{-1} x \implies x = \sin \theta, \ \sqrt{1-x^2} = \cos \theta

32x+121−x2=32sin⁡θ+12cos⁡θ=cos⁡(π3−θ)\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} = \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \cos\left(\frac{\pi}{3}-\theta\right)

⇒sin⁡−1(32x+121−x2)=sin⁡−1(cos⁡(π3−θ))\Rightarrow \sin^{-1}\left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) = \sin^{-1} \left( \cos(\frac{\pi}{3}-\theta) \right)

=π2−(π3−θ)=π6+θ=π6+sin⁡−1x= \frac{\pi}{2} - (\frac{\pi}{3}-\theta) = \frac{\pi}{6} + \theta = \frac{\pi}{6} + \sin^{-1} x

Answer key and solution verified before publishing.

Practise Inverse Trigonometric Functions

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions
Considering the principal values of the inverse trigonometric… | JEE Main 2025 PYQ with Solution · DhiX AI