If sin inverse x + sin inverse y + sin inverse z = pie, prove that x square - y - Maths - Inverse Trigonometric Functions

Question

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

Rupak Bhattacharya · Accepted Answer

sin-1x+sin-1y+sin-1z=πLet sin-1x = A; sin-1y = B; sin-1z = C⇒x = 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=π⇒A + B + C = π⇒B+C = π-A⇒cosB+C = cosπ-A⇒cos B 
 cos C - sin B 
 sin C = - cos A⇒1-y21-z2 - yz = -1-x2⇒ 1-y21-z2 = yz - 1-x2⇒1-y21-z2 = y2z2+1-x2-2yz1-x2⇒x2-y2-z2+2yz1-x2 = 0

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