1.

If `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi`, show that `x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+z^2x^2)`

Answer» `sin^-1x+sin^-1y +sin^-1z = pi`
`=> sin^-1x+sin^-1y = pi - sin^-1z`
`=> sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2)) = pi-sin^-1z`
Taking sin both sides,
`=> sin(sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))) = sin(pi- sin^-1z)`
`=> xsqrt(1-y^2)+ysqrt(1-x^2) =z`
`=> xsqrt(1-y^2) =z - ysqrt(1-x^2)`
Taking square both sides,
`=>x^2(1-y^2) = z^2+y^2(1-x^2)-2yzsqrt(1-x^2)`
`=>x^2-x^2y^2 = z^2+y^2-x^2y^2 - 2yzsqrt(1-x^2)`
`=>x^2-y^2-z^2 = -2yzsqrt(1-x^2)`
Again squaring both sides,
`=>(x^2-y^2-z^2)^2 = 4y^2z^2(1-x^2)`
`=>x^2+y^4+z^4+2y^2z^2-2x^2y^2-2x^2z^2 = 4y^2z^2-4x^2y^2z^2`
`=>x^2+y^4+z^4+4x^2y^2z^2 = 2(x^2y^2+y^2z^2+z^2x^2)`


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