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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

Considering the principal values of the inverse trigonometric...

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

Options

  1. Aπ4+sin⁡−1x\frac{\pi}{4}+\sin ^{-1} x
  2. Bπ6+sin⁡−1x\frac{\pi}{6}+\sin ^{-1} x
  3. C−5π6−sin⁡−1x\frac{-5 \pi}{6}-\sin ^{-1} x
  4. D5π6−sin⁡−1x\frac{5 \pi}{6}-\sin ^{-1} x

Correct answer

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

Step-by-step solution

sin⁡−1(32x+121−x2),−12<x<12⇒ Let sin⁡−1(x)=θ−π6<θ<π4⇒x=sin⁡θ, then ⇒sin⁡−1(32sin⁡θ+12cos⁡θ)⇒sin⁡−1(sin⁡(θ+π6))=θ+π6⇒sin⁡−1(x)+π6\begin{aligned} & \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right), \frac{-1}{2} < x < \frac{1}{\sqrt{2}} \\ & \Rightarrow \text{ Let } \sin ^{-1}(x)=\theta \quad \frac{-\pi}{6} < \theta < \frac{\pi}{4} \\ & \Rightarrow x=\sin \theta, \text{ then } \\ & \Rightarrow \sin ^{-1}\left(\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta\right) \\ & \Rightarrow \sin ^{-1}\left(\sin \left(\theta+\frac{\pi}{6}\right)\right)=\theta+\frac{\pi}{6} \\ & \Rightarrow \sin ^{-1}(x)+\frac{\pi}{6}\end{aligned}
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