2sin−1⁡x+3cos−1⁡y=α,3cos−1⁡x+2sin−1⁡y =β. If α+β=7π3, then xy−1−x21−y2 is equal to

7π3=α+β=2×π2+cos−1⁡x+2×π2+cos−1⁡y⇒cos−1⁡x+cos−1⁡y=7π3−2π=π3⇒cos⁡cos−1⁡x+cos−1⁡y=cos⁡(π/3)=1/2⇒xy−1−x21−y2=1/2

2sin−1⁡x+3cos−1⁡y=α,3cos−1⁡x+2sin−1⁡y =β. If α+β=7π3, then xy−1−x21−y2 is equal to

7π3=α+β=2×π2+cos−1⁡x+2×π2+cos−1⁡y⇒cos−1⁡x+cos−1⁡y=7π3−2π=π3⇒cos⁡cos−1⁡x+cos−1⁡y=cos⁡(π/3)=1/2⇒xy−1−x21−y2=1/2

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$2 \sin^{- 1} x + 3 \cos^{- 1} y = α, 3 \cos^{- 1} x + 2 \sin^{- 1} y$ $= β . If α + β = \frac{7 π}{3}, then x y - \sqrt{(1 - x^{2}) (1 - y^{2})}$ is equal to

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$1 / 2$

$\sqrt{3} / 2$

$1 / \sqrt{2}$

answer is B.

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$\frac{7 π}{3} = α + β = 2 \times \frac{π}{2} + \cos^{- 1} x + 2 \times \frac{π}{2} + \cos^{- 1} y$

$\begin{array}{l} \Rightarrow \cos^{- 1} x + \cos^{- 1} y = \frac{7 π}{3} - 2 π = \frac{π}{3} \\ \Rightarrow \cos (\cos^{- 1} x + \cos^{- 1} y) = \cos (π / 3) = 1 / 2 \\ \Rightarrow x y - \sqrt{(1 - x^{2}) (1 - y^{2})} = 1 / 2 \end{array}$

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2sin−1⁡x+3cos−1⁡y=α,3cos−1⁡x+2sin−1⁡y =β. If α+β=7π3, then xy−1−x21−y2 is equal to

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$2 \sin^{- 1} x + 3 \cos^{- 1} y = α, 3 \cos^{- 1} x + 2 \sin^{- 1} y$ $= β . If α + β = \frac{7 π}{3}, then x y - \sqrt{(1 - x^{2}) (1 - y^{2})}$ is equal to