If `cos^(-1)x+cos^(-1)y+cos^(-1)z=pi,t h e n` `x^2+y^2+z^2+2x y z=1` `2(sin^(-1)x+sin^(-1)y+sin^(-1)z)=cos^(-1)x+cos^(-1)y+cos^(-1)z` `x y+y z+z x=x+y+z-1` `(x+1/x)+(y+1/y)+(z+1/z)geq6`

Explanation

`cos^-1x+cos^-1y+cos^-1z = pi`

`=>pi/2-sin^-1x+pi/2-sin^-1y+pi/2-sin^-1z = pi`

`=>sin^-1x+sin^-1y+sin^-1z = pi/2`

`=>2(sin^-1x+sin^-1y+sin^-1z) = pi`

`=>2(sin^-1x+sin^-1y+sin^-1z) = cos^-1x+cos^-1y+cos^-1z`

So, option -`b` is correct.

Now,

`cos^-1x+cos^-1y+cos^-1z = pi`

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