If sin inverse x + sin inverse y + sin inverse z = pie, prove that x square - y square - z square + 2yz (square root of (1-x square)=0

sin-1x+sin-1y+sin-1z=πLet sin-1x = A; sin-1y = B; sin-1z = Cx = sin A; y = sin B; z = sin CNow, cosA = 1-sin2A = 1-x2cos B = 1-sin2B = 1-y2cos C = 1-sin2C = 1-z2Now, sin-1x+sin-1y+sin-1z=πA + B + C = πB+C = π-AcosB+C = cosπ-Acos B . cos C - sin B . sin C = - cos A1-y21-z2 - yz = -1-x2 1-y21-z2 = yz - 1-x21-y21-z2 = y2z2+1-x2-2yz1-x2x2-y2-z2+2yz1-x2 = 0

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