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Question 53 of 71

Q.If sin⁡−1x+sin⁡−1y=2π3\sin^{-1}x + \sin^{-1}y = \dfrac{2\pi}{3}, then cos⁡−1x+cos⁡−1y\cos^{-1}x + \cos^{-1}y is equal to :

(a) π\pi
(b) 2π3\dfrac{2\pi}{3}
(c) π3\dfrac{\pi}{3}
(d) π6\dfrac{\pi}{6}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2020MCQ· 1mImportance★★★★★
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Using the identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x+\cos^{-1}x=\pi/2 applied to both xx and yy, cos⁡−1x+cos⁡−1y=π−2π3=π3\cos^{-1}x+\cos^{-1}y=\pi-\dfrac{2\pi}{3}=\dfrac{\pi}{3}.

  1. Recall the standard identity: for any x∈[−1,1]x\in[-1,1], sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}.
  2. Apply this identity to xx: sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}.
  3. Apply the same identity to yy: sin⁡−1y+cos⁡−1y=π2\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}.
  4. Add these two equations: (sin⁡−1x+sin⁡−1y)+(cos⁡−1x+cos⁡−1y)=π2+π2=π\left(\sin^{-1}x+\sin^{-1}y\right)+\left(\cos^{-1}x+\cos^{-1}y\right)=\dfrac{\pi}{2}+\dfrac{\pi}{2}=\pi. …

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