let 0ltphiltpi/2, x=sum(n=0)^oocos^(2n)p...

let `0ltphiltpi/2`, `x=sum_(n=0)^oocos^(2n)phi`, `y=sum_(n=0)^oosin^(2n)phi` and `z=sum_(n=0)^oocos^(2n)phisin^(2n)phi`

`xyz = xz+y`

`xyz = xy+z`

`xyz=x+y+z`

`xyz =yz +x`

Text Solution

Verified by Experts

The correct Answer is:

Since , `x = sum _(n=0)^(oo) cos^(2n) phi `
`1 + cos^(2) phi + cos^(4) phi +….`
`= 1/(1-cos^(2) phi)=1/(sin^(2)phi)" " [ :' | cos x |lt 1]`
Similarly ` y = 1/(1-sin^(2)phi )= 1/(sin^(2)phi )`
and ` z = 1/(1-sin^(2)phi cos^(2)phi)`
` = 1/(1-1/x*1/y)=(xy)/(xy-1)`
` rArr xyz = xy +z`

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If 0 lt theta, phi lt (pi)/(2), x = sum_(n=0)^(oo)cos^(2n)theta, y=sum_(n=0)^(oo)sin^(2n)phi and z=sum_(n=0)^(oo)cos^(2n)theta*sin^(2n)phi then :

`xy-z=(x+y)z`

`xy+yz+zx=z`

`xyz=4`

`xy+z=(x+y)z`

For 0 lt phi le ( pi )/( 2) , if x = sum_(n=0)^(oo) cos^(2n) phi , y = sum_(n=0)^(oo) sin ^(2n) phi , z= sum _(n=0)^(oo) cos^(2n) phi sin ^(2n ) phi , then

`xyz = xz +y`

`xyz = xy +z`

`xyz = x+y+z`

`xyz = yz + x`

If x=sum_(n=0)^(oo) a^(n),y=sum_(n=0)^(oo)b^(n),z=sum_(n=0)^(oo)(ab)^(n) , where a,blt1 , then

xyz=x+y+z

xz+yz=xy+z

xy+yz=xz+y

xy+xz=yz+x