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Inverse Trigonometric Functions question

2021 · 27 Aug · Shift 1 · Q24
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  5. /2021 · 27 Aug · Shift 1 · Q24

Inverse Trigonometric Functions question

2021 · 27 Aug · Shift 1 · Q24

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If (sin⁡−1x)2−(cos⁡−1x)2=a{({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a(sin−1x)2−(cos−1x)2=a; 0 < x < 1, a eee 0, then the value of 2x2 −-− 1 is :
  1. A
    cos⁡(4aπ)\cos \left( {{{4a} \over \pi }} \right)cos(π4a​)
  2. B
    sin⁡(2aπ)\sin \left( {{{2a} \over \pi }} \right)sin(π2a​)
  3. C
    cos⁡(2aπ)\cos \left( {{{2a} \over \pi }} \right)cos(π2a​)
  4. D
    sin⁡(4aπ)\sin \left( {{{4a} \over \pi }} \right)sin(π4a​)
View written solutionFree

Correct answer: B

  1. Let α=sin⁡−1x,β=cos⁡−1x\alpha = \sin^{-1}x, \qquad \beta = \cos^{-1}xα=sin−1x,β=cos−1x for 0<x<10<x<10<x<1.

Then we know the standard identity: α+β=π2\alpha + \beta = \frac{\pi}{2}α+β=2π​

  1. Given: (sin⁡−1x)2−(cos⁡−1x)2=a({\sin^{-1}x})^2-({\cos^{-1}x})^2=a(sin−1x)2−(cos−1x)2=a so α2−β2=a\alpha^2-\beta^2=aα2−β2=a

Using difference of squares: α2−β2=(α−β)(α+β)\alpha^2-\beta^2=(\alpha-\beta)(\alpha+\beta)α2−β2=(α−β)(α+β) Hence, a=(α−β)⋅π2a=(\alpha-\beta)\cdot \frac{\pi}{2}a=(α−β)⋅2π​ which gives α−β=2aπ\alpha-\beta=\frac{2a}{\pi}α−β=π2a​

  1. Now solve for α\alphaα and β\betaβ using α+β=π2,α−β=2aπ\alpha+\beta=\frac{\pi}{2}, \qquad \alpha-\beta=\frac{2a}{\pi}α+β=2π​,α−β=π2a​ Adding, 2α=π2+2aπ2\alpha=\frac{\pi}{2}+\frac{2a}{\pi}2α=2π​+π2a​ so α=π4+aπ\alpha=\frac{\pi}{4}+\frac{a}{\pi}α=4π​+πa​

  2. Since x=sin⁡αx=\sin\alphax=sinα, we need 2x2−1=2sin⁡2α−1=−cos⁡2α2x^2-1=2\sin^2\alpha-1=-\cos 2\alpha2x2−1=2sin2α−1=−cos2α Now, 2α=π2+2aπ2\alpha=\frac\pi2+\frac{2a}{\pi}2α=2π​+π2a​ Therefore, 2x2−1=−cos⁡(π2+2aπ)2x^2-1=-\cos\left(\frac\pi2+\frac{2a}{\pi}\right)2x2−1=−cos(2π​+π2a​) Using cos⁡(π2+θ)=−sin⁡θ,\cos\left(\frac\pi2+\theta\right)=-\sin\theta,cos(2π​+θ)=−sinθ, we get 2x2−1=−(−sin⁡2aπ)=sin⁡(2aπ)2x^2-1=-(-\sin\tfrac{2a}{\pi})=\sin\left(\frac{2a}{\pi}\right)2x2−1=−(−sinπ2a​)=sin(π2a​)

  3. Hence the correct option is sin⁡(2aπ)\boxed{\sin\left(\frac{2a}{\pi}\right)}sin(π2a​)​ which is Option B.

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