, then is equal…

MATHEMATICS · Single correct · CBSE · Mathematics · Inverse Trigonometric Functions · Inverse Trigonometric Functions

Problem

(sin⁡−1x)2+(sin⁡−1y)2+2(sin⁡−1x)(sin⁡−1y)=π2(\sin^{-1} x)^2 + (\sin^{-1} y)^2 + 2(\sin^{-1} x)(\sin^{-1} y) = \pi^2 , then x2+y2x^2 + y^2 is equal to: (a) 1 (b) 3/23/2 (c) 2 (d) 1/21/2

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