sin^-1x+sin^-1y+sin^-1z=pi/2 then find x^2+ y^2+z^2+ 2xyz =?

sin-¹(x)+sin-¹(y)+sin-¹(z)=π/2Using sin-¹(a)=π/2-cos-¹(a)π/2-cos-¹(x)+π/2-cos-¹(y)+π/2-cos-¹(z)=π/2cos-¹(x)+cos-¹(y)+cos-¹(z)=πcos-¹(xy-√(1-x²)√(1-y²))=π-cos-¹(z)=cos-¹(-z)=> xy-√(1-x²)(1-y²)=-z√(1-x²)(1-y²)=z+xySquaring both sides1-y²-x²+x²y²=z²+x²y²+2xyzx²+y²+z²+2xyz=1

Answers : (1)sin-¹(x)+sin-¹(y)+sin-¹(z)=π/2 Using sin-¹(a)=π/2-cos-¹(a) π/2-cos-¹(x)+π/2-cos-¹(y)+π/2-cos-¹(z)=π/2 cos-¹(x)+cos-¹(y)+cos-¹(z)=π cos-¹(xy-√(1-x²)√(1-y²))=π-cos-¹(z)=cos-¹(-z) => xy-√(1-x²)(1-y²)=-z √(1-x²)(1-y²)=z+xy Squaring both sides 1-y²-x²+x²y²=z²+x²y²+2xyz x²+y²+z²+2xyz=1

Answers : (1)sin-¹(x)+sin-¹(y)+sin-¹(z)=π/2. Using sin-¹(a)=π/2-cos-¹(a). π/2-cos-¹(x)+π/2-cos-¹(y)+π/2-cos-¹(z)=π/2. cos-¹(x)+cos-¹(y)+cos-¹(z)=π. cos-¹(xy-√(1-x²)√(1-y²))=π-cos-¹(z)=cos-¹(-z). => xy-√(1-x²)(1-y²)=-z. √(1-x²)(1-y²)=z+xy. Squaring both sides. 1-y²-x²+x²y²=z²+x²y²+2xyz. x²+y²+z²+2xyz=1