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Inverse Trigonometric Functions question

2024 · 31 Jan · Shift 1 · Q32
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  5. /2024 · 31 Jan · Shift 1 · Q32

Inverse Trigonometric Functions question

2024 · 31 Jan · Shift 1 · Q32

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
For α,β,γeq0\alpha, \beta, \gamma eq 0α,β,γeq0, if sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pisin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta(α+β+γ)(α−γ+β)=3αβ, then γ\gammaγ equals
  1. A
    3\sqrt{3}3​
  2. B
    32\frac{\sqrt{3}}{2}23​​
  3. C
    12\frac{1}{\sqrt{2}}2​1​
  4. D
    3−122\frac{\sqrt{3}-1}{2 \sqrt{2}}22​3​−1​
View written solutionFree

Correct answer: B

  1. Let A=sin⁡−1α,B=sin⁡−1β,C=sin⁡−1γ.A=\sin^{-1}\alpha,\quad B=\sin^{-1}\beta,\quad C=\sin^{-1}\gamma.A=sin−1α,B=sin−1β,C=sin−1γ. Then A+B+C=π,A+B+C=\pi,A+B+C=π, with α=sin⁡A, β=sin⁡B, γ=sin⁡C.\alpha=\sin A,\ \beta=\sin B,\ \gamma=\sin C.α=sinA, β=sinB, γ=sinC.

  2. Since A+B=π−CA+B=\pi-CA+B=π−C, we have sin⁡(A+B)=sin⁡C=γ.\sin(A+B)=\sin C=\gamma.sin(A+B)=sinC=γ. Expanding: sin⁡Acos⁡B+cos⁡Asin⁡B=γ,\sin A\cos B+\cos A\sin B=\gamma,sinAcosB+cosAsinB=γ, i.e. α1−β2+β1−α2=γ.\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}=\gamma. α1−β2​+β1−α2​=γ. This is not directly needed; instead we use a standard identity for three angles summing to π\piπ:

    If A+B+C=πA+B+C=\piA+B+C=π, then sin⁡2A+sin⁡2B+sin⁡2C=2sin⁡Asin⁡Bsin⁡C+2cos⁡Acos⁡Bcos⁡C.\sin^2 A+\sin^2 B+\sin^2 C=2\sin A\sin B\sin C+2\cos A\cos B\cos C.sin2A+sin2B+sin2C=2sinAsinBsinC+2cosAcosBcosC. But here a more useful relation comes from cos⁡(A+B)=−cos⁡C.\cos(A+B)= -\cos C.cos(A+B)=−cosC. So cos⁡Acos⁡B−sin⁡Asin⁡B=−1−γ2.\cos A\cos B-\sin A\sin B=-\sqrt{1-\gamma^2}.cosAcosB−sinAsinB=−1−γ2​.

  3. Now use the given algebraic condition: (α+β+γ)(α−γ+β)=3αβ. (\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3\alpha\beta.(α+β+γ)(α−γ+β)=3αβ. Rewrite the second factor: α−γ+β=α+β−γ.\alpha-\gamma+\beta=\alpha+\beta-\gamma.α−γ+β=α+β−γ. Hence [(α+β)+γ][(α+β)−γ]=3αβ,[(\alpha+\beta)+\gamma][(\alpha+\beta)-\gamma]=3\alpha\beta,[(α+β)+γ][(α+β)−γ]=3αβ, so (α+β)2−γ2=3αβ.(\alpha+\beta)^2-\gamma^2=3\alpha\beta.(α+β)2−γ2=3αβ. Expanding, α2+β2+2αβ−γ2=3αβ,\alpha^2+\beta^2+2\alpha\beta-\gamma^2=3\alpha\beta,α2+β2+2αβ−γ2=3αβ, therefore α2+β2−αβ=γ2.(1)\alpha^2+\beta^2-\alpha\beta=\gamma^2. \qquad (1)α2+β2−αβ=γ2.(1)

  4. Since A+B=π−CA+B=\pi-CA+B=π−C, take sine squared: γ2=sin⁡2(A+B).\gamma^2=\sin^2(A+B).γ2=sin2(A+B). Now sin⁡2(A+B)=sin⁡2Acos⁡2B+sin⁡2Bcos⁡2A+2sin⁡Asin⁡Bcos⁡Acos⁡B.\sin^2(A+B)=\sin^2 A\cos^2 B+\sin^2 B\cos^2 A+2\sin A\sin B\cos A\cos B.sin2(A+B)=sin2Acos2B+sin2Bcos2A+2sinAsinBcosAcosB. Using sin⁡A=α, sin⁡B=β\sin A=\alpha,\ \sin B=\betasinA=α, sinB=β, this becomes γ2=α2(1−β2)+β2(1−α2)+2αβcos⁡Acos⁡B.\gamma^2=\alpha^2(1-\beta^2)+\beta^2(1-\alpha^2)+2\alpha\beta\cos A\cos B.γ2=α2(1−β2)+β2(1−α2)+2αβcosAcosB. So γ2=α2+β2−2α2β2+2αβcos⁡Acos⁡B.(2)\gamma^2=\alpha^2+\beta^2-2\alpha^2\beta^2+2\alpha\beta\cos A\cos B. \qquad (2)γ2=α2+β2−2α2β2+2αβcosAcosB.(2)

  5. Compare (1) and (2): α2+β2−αβ=α2+β2−2α2β2+2αβcos⁡Acos⁡B.\alpha^2+\beta^2-\alpha\beta=\alpha^2+\beta^2-2\alpha^2\beta^2+2\alpha\beta\cos A\cos B.α2+β2−αβ=α2+β2−2α2β2+2αβcosAcosB. Hence −αβ=−2α2β2+2αβcos⁡Acos⁡B.-\alpha\beta=-2\alpha^2\beta^2+2\alpha\beta\cos A\cos B.−αβ=−2α2β2+2αβcosAcosB. Since α,β≠0\alpha,\beta\ne 0α,β=0, divide by αβ\alpha\betaαβ: −1=−2αβ+2cos⁡Acos⁡B,-1=-2\alpha\beta+2\cos A\cos B,−1=−2αβ+2cosAcosB, i.e. 2cos⁡Acos⁡B=2αβ−1.(3)2\cos A\cos B=2\alpha\beta-1. \qquad (3)2cosAcosB=2αβ−1.(3)

  6. But from cos⁡(A+B)=cos⁡(π−C)=−cos⁡C=−1−γ2,\cos(A+B)=\cos(\pi-C)=-\cos C=-\sqrt{1-\gamma^2},cos(A+B)=cos(π−C)=−cosC=−1−γ2​, we get cos⁡Acos⁡B−αβ=−1−γ2.\cos A\cos B-\alpha\beta=-\sqrt{1-\gamma^2}.cosAcosB−αβ=−1−γ2​. Using (3), cos⁡Acos⁡B=αβ−12.\cos A\cos B=\alpha\beta-\frac12.cosAcosB=αβ−21​. Therefore αβ−12−αβ=−1−γ2,\alpha\beta-\frac12-\alpha\beta=-\sqrt{1-\gamma^2},αβ−21​−αβ=−1−γ2​, so −12=−1−γ2.-\frac12=-\sqrt{1-\gamma^2}.−21​=−1−γ2​. Thus 1−γ2=12.\sqrt{1-\gamma^2}=\frac12.1−γ2​=21​.

  7. Squaring: 1−γ2=141-\gamma^2=\frac141−γ2=41​ γ2=34\gamma^2=\frac34γ2=43​ Since γ=sin⁡C\gamma=\sin Cγ=sinC and C∈[−π/2,π/2]C\in[-\pi/2,\pi/2]C∈[−π/2,π/2], here from the sum being π\piπ we must have positive values, so γ>0\gamma>0γ>0. Hence γ=32.\gamma=\frac{\sqrt3}{2}. γ=23​​.

  8. Option check:

    • A: 3\sqrt33​ is impossible since ∣γ∣≤1|\gamma|\le 1∣γ∣≤1.
    • B: 32\dfrac{\sqrt3}{2}23​​ ✔
    • C: 12\dfrac1{\sqrt2}2​1​ ✘
    • D: 3−122\dfrac{\sqrt3-1}{2\sqrt2}22​3​−1​ ✘

Therefore the correct answer is B.

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