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If 0 lt phi lt (pi)/(2) , x= sum(n=0)...

If `0 lt phi lt (pi)/(2) , x= sum_(n=0)^(oo) cos^(2n) phi ,y sum _(n=0)^(oo) sin ^(2n) phi and z = sum _(n=0) ^(oo) cos ^(2n) phi sin ^(2n) phi , ` then

A

`xyz=xz+y`

B

`xyz=xy+z`

C

`xyz=x+y+z`

D

`xyz=yz+x`

Text Solution

Verified by Experts

The correct Answer is:
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)]
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