For `0 lt phi lt pi//2 , if x =Sigma_(n=0)^(oo) cos^(2n) phi ,y=Sigma_(n=0)^(oo) sin ^(2n)phi` and `z=Sigma_(n=0)^(oo) cos^(2n)phi then`

Options

A. xyz=xz+y
B. xyz=xy +z
C. xyz = z+y+z
D. xyz =yz +x

Answer

B, C

Explanation

We have for 0`ltphipi//2`,
`x=sum_(n=0)^(oo)cos^(2n)phi=1+cos^(2)phi+cos^(4)phi+..oo=1/(1-cos^(2)phi)`
`rArrx=1/(sin^(2)phi)` …..(1)
`y=sum_(n=0)^(oo)sin^(2n)phi=1+sin^(2)phi+sin^(4)phi+..oo=1/(1-sin^(2)phi)`
`rArry=1/(cos^(2)phi)`
`z=sum_(n=0)^(oo)cos^(2n)phisin^(2n)phi`
`=1+cos^(2)phisin^(2)phi+cos^(4)phisin^(4)phi+...oo`
`rArrz=1/(1-cos^(2)phisin^(2)phi)` ...(3)
Substituting the values of `cos^(2)phi and sin^(2)phi` in (3), from (1) and (2), we get
`z=1/(1-1/xx1/y)`
`rArrz=(xy)/(xy-1)`
`rArrxyz-z=xy`
`rArrxyz=xy+z`
Also, `x+y+z=1/(sin^(2)phi)+1/(cos^(2)phi)+1/(1-cos^(2)phisin^(2)phi)`
`cos^(2)phi(1-cos^(2)phisin^(2)phi)+sin^(2)phi(1-cos^(2)phisin^(2)phi)+(cos^(2)phisin^(2)phi)/(cos^(2)phisin^(2)phi(1-cos^(2)phisin^(2)phi))`
`=((sin^(2)phi+cos^(2)phi)(1-cos^(2)phisin^(2)phi)+cos^(2)phisin^(2)phi)/(cos^(2)phisin^(2)phi(1-cos^(2)sin^(2)phi))`
`1/(cos^(2)phisin^(2)phi(1-cos^(2)phisin^(2)phi))=xyz`