sin−1x+sin−1y+sin−1z=πLet x=sina,y=sinb and z=sinc∴a+b+c=πx√1−x2+y√1−y2+z√1−z2=sina√1−sin2a+sinb√1−sin2b+sinc√1−sin2c=sinacosa+sinbcosb+sinccosc=sin2a+sin2b+sin2c2=2sin2a+2b2cos2a−2b2+sin2c2=2sin a+b cos a−b sin2c2=2sin(π−c)cos(a−b)+2sinc cosc2=2sinc cos(a−b)+2sinc cosc2=sinc[cos(a−b)+cosc]=sinc[2cos(a−b+c2)cos(a−b−c2)]=sinc[2cosπ−b−b2cos(a−π+a2)]=sinc[2cos(π2−b)cos(a−π2)]=sinc×2×sinb×sina=2sina sinb sinc=2xyz