Correct Answer - B::C
All are infinte geometric progression with common ratio lt 1
`x=1/(1-cos^2phi)=1/sin^2phi,y=1/(1-sin^2phi)=1/cos^2phi`,
`z=1/(1-cos^2phisin^2phi)`
Now, `xy+z=1/(sin^2phicos^2phi)+1/(1-sin^2phicos^2phi)`
`=1/(sin^2phicos^2phi(1-sin^2phicos^2phi))`
`or xy+z=xyz ...(i)`
Clearly, `x+y=(sin^2phi+cos^2phi)/(sin^2phicos^2phi)=xy`
`:. x+y+z=xyz` [using Eq. (i)]