If `cos^(-1)x + cos^(-1)y + cos^(-1)z = 3pi,` then `xy + yz +zx` is equal to

Options

A. `x^(2) + y^(2) + z^(2) + 2xyz = 1`
B. `2(sin^(-1) x + sin^(-1) y + sin^(-1) z) = cos^(-1) x + cos^(-1) y + cos^(-1) z`
C. `xy + yz + zx = x + y + z -1`
D. `(x + (1)/(x)) + (y + (1)/(y)) + (z + (1)/(z)) gt 6`

Answer

A, B

Explanation

`cos^(-1)x + cos^(-1) y + cos^(-1) z = pi`
or `sin^(-1)x + sin^(-1) y + sin^(-1) z = pi//2`
or `cos^(-1) x + cos^(-1) y = cos^(-1) (-z)`
`rArr xy - sqrt(1 -x^(2)) sqrt(1 -y^(2)) = -z`
or `x^(2) + y^(2) + z^(2) + 2xyz = 1`