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Inverse Trigonometric Functions question

2025 · 4 Apr · Shift 1 · Q45
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  5. /2025 · 4 Apr · Shift 1 · Q45

Inverse Trigonometric Functions question

2025 · 4 Apr · Shift 1 · Q45

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
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}}sin−1(23​​x+21​1−x2​),−21​<x<2​1​, is equal to
  1. A
    −5π6−sin⁡−1x\frac{-5 \pi}{6}-\sin ^{-1} x6−5π​−sin−1x
  2. B
    5π6−sin⁡−1x\frac{5 \pi}{6}-\sin ^{-1} x65π​−sin−1x
  3. C
    π6+sin⁡−1x\frac{\pi}{6}+\sin ^{-1} x6π​+sin−1x
  4. D
    π4+sin⁡−1x\frac{\pi}{4}+\sin ^{-1} x4π​+sin−1x
View written solutionFree

Correct answer: C

  1. Let θ=sin⁡−1x.\theta=\sin^{-1}x.θ=sin−1x. Since −12<x<12-\dfrac12<x<\dfrac1{\sqrt2}−21​<x<2​1​ and θ\thetaθ is the principal value of sin⁡−1x\sin^{-1}xsin−1x, we have θ∈(−π6,π4).\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right).θ∈(−6π​,4π​).

  2. Then x=sin⁡θ,1−x2=1−sin⁡2θ=cos⁡θ.x=\sin\theta,\qquad \sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta.x=sinθ,1−x2​=1−sin2θ​=cosθ. This is valid because for principal values of sin⁡−1\sin^{-1}sin−1, θ∈[−π2,π2]\theta\in\left[-\frac\pi2,\frac\pi2\right]θ∈[−2π​,2π​], so cos⁡θ≥0\cos\theta\ge 0cosθ≥0.

  3. Substitute into the given expression:

    =\frac{\sqrt3}{2}\sin\theta+\frac12\cos\theta.$$

  4. Recognize the sine addition formula:

    =\frac{\sqrt3}{2}\sin\theta+\frac12\cos\theta.$$ Hence, $$\frac{\sqrt3}{2}x+\frac12\sqrt{1-x^2}=\sin\left(\theta+\frac\pi6\right).$$

  5. Therefore the given expression becomes sin⁡−1(sin⁡(θ+π6)).\sin^{-1}\left(\sin\left(\theta+\frac\pi6\right)\right).sin−1(sin(θ+6π​)).

  6. Now check the range of θ+π6\theta+\dfrac\pi6θ+6π​:

    \implies \theta+\frac\pi6\in\left(0,\frac{5\pi}{12}\right).$$ Since $$\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac\pi2,\frac\pi2\right],$$ this lies entirely in the principal range of $\sin^{-1}$. So, $$\sin^{-1}(\sin y)=y \quad \text{for } y\in\left[-\frac\pi2,\frac\pi2\right].$$

  7. Hence,

    =\theta+\frac\pi6 =\sin^{-1}x+\frac\pi6.$$

  8. So the correct option is π6+sin⁡−1x.\boxed{\frac\pi6+\sin^{-1}x}.6π​+sin−1x​.

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