JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

If $0<x<\dfrac1{\sqrt2}$ and $\dfrac{\sin^{-1}x}{\alpha}=\dfrac{\cos^{-1}x}{\beta}$ , then a value of $\sin\left(\dfrac{2\pi\alpha}{\alpha+\beta}\right)$ is

4\sqrt{\left(1-x^2\right)}\left(1-2x^2\right)

4x\sqrt{\left(1-x^2\right)}\left(1-2x^2\right)

2x\sqrt{\left(1-x^2\right)}\left(1-4x^2\right)

4\sqrt{\left(1-x^2\right)}\left(1-4x^2\right)

Answer and worked solution

Official answer: B

Common ratio. Each ratio equals $\frac{\sin^{-1}x+\cos^{-1}x}{\alpha+\beta}=\frac{\pi/2}{\alpha+\beta}$ , so $\frac{2\pi\alpha}{\alpha+\beta}=4\sin^{-1}x$ .

With $\sin\theta=x$ : $\sin4\theta=2\sin2\theta\cos2\theta=4x\sqrt{1-x^2}\,(1-2x^2)$ .

(B)

Topics:Inverse trigonometric functions Trigonometric ratios and identities

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 16