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For 0ltphiltpi//2," if" x=sum(n=0)^(oo) ...

For `0ltphiltpi//2," if" x=sum_(n=0)^(oo) cos^(2n)phi,y=sum_(n=0)^(oo) sin^(2m)phi,z=sum_(n=0)^(oo)cos^(2n)phisin^(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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