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MCQMediumJEE 2025Inverse Trigonometric Functions

JEE Mathematics 2025 Question with Solution

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<x<12-\frac{1}{2} < x < \frac{1}{\sqrt{2}}, is equal to

  • A

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

  • B

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

  • C

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

  • D

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: sin⁡−1(32x+121−x2)\sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) with −12<x<12-\frac{1}{2} < x < \frac{1}{\sqrt{2}}.

Find: Its principal value.

Let

θ=sin⁡−1x\theta = \sin^{-1} x

Then

x=sin⁡θx = \sin \theta

and since θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], we have

1−x2=cos⁡θ\sqrt{1-x^2} = \cos \theta

Principal Value Check

Substitute in the given expression:

sin⁡−1(32sin⁡θ+12cos⁡θ)\sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right)

Using

cos⁡π6=32,sin⁡π6=12\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \qquad \sin \frac{\pi}{6} = \frac{1}{2}

this becomes

sin⁡−1(sin⁡θcos⁡π6+cos⁡θsin⁡π6)\sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) =sin⁡−1(sin⁡(θ+π6))= \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right)

Using the interval directly

From −12<x<12-\frac{1}{2} < x < \frac{1}{\sqrt{2}}, we get

θ=sin⁡−1x∈(−π6,π4)\theta = \sin^{-1}x \in \left(-\frac{\pi}{6}, \frac{\pi}{4}\right)

Hence

θ+π6∈(0,5π12)\theta + \frac{\pi}{6} \in \left(0, \frac{5\pi}{12}\right)

which lies completely inside the principal range of sin⁡−1\sin^{-1}, namely [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Therefore,

sin⁡−1(sin⁡(θ+π6))=θ+π6\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)=\theta+\frac{\pi}{6}

So,

sin⁡−1(32x+121−x2)=sin⁡−1x+π6\sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right)=\sin^{-1}x+\frac{\pi}{6}

Therefore, the correct option is B.

Common mistakes

  • Taking sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y)=y for every real yy is incorrect. This is valid only when yy lies in the principal range [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. First verify the interval of θ+π6\theta+\frac{\pi}{6}.

  • Using 1−x2=±cos⁡θ\sqrt{1-x^2}=\pm \cos\theta without fixing θ=sin⁡−1x\theta=\sin^{-1}x is wrong. Since θ\theta is in the principal range of inverse sine, cos⁡θ≥0\cos\theta\ge 0, so 1−x2=cos⁡θ\sqrt{1-x^2}=\cos\theta.

  • Applying the angle addition identity incorrectly is a common error. Here 32=cos⁡π6\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6} and 12=sin⁡π6\frac{1}{2}=\sin\frac{\pi}{6}, so the expression becomes sin⁡(θ+π6)\sin\left(\theta+\frac{\pi}{6}\right), not sin⁡(θ−π6)\sin\left(\theta-\frac{\pi}{6}\right).

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