If `x=Sigma_(n=0)^(oo) cos^(2n) theta, y= Sigma_(n=0)^(oo) sin^(2n) phi` and `z= Sigma_(n=0)^(oo) cos^(2n) theta sin^(2n) phi`, where `0 lt theta lt phi lt pi/2` then prove that `xz+yz-z=xy`.

If `x=Sigma_(n=0)^(oo) cos^(2n) theta, y= Sigma_(n=0)^(oo) sin^(2n) ph

1.	If `x=Sigma_(n=0)^(oo) cos^(2n) theta, y= Sigma_(n=0)^(oo) sin^(2n) phi` and `z= Sigma_(n=0)^(oo) cos^(2n) theta sin^(2n) phi`, where `0 lt theta lt phi lt pi/2` then prove that `xz+yz-z=xy`.
Answer» We have `x=sum_(n=0)^(oo) cos^(2n)theta=1+cos^(2) theta+ cos^(4) theta+...oo` `= a/((1-r))`, where `a=1` and `r= cos^(2) theta` [sum of an infinite GP] `=1/((1- cos^(2) theta))=1/(sin^(2) theta)`. `:. sin^(2) theta=1/x` ...(i) `y= sum_(n=0)^(oo) sin^(2n) phi=1 + sin^(2) phi + sin^(4) phi +...oo` `=a/((1-r))`, where `a=1` and `r= sin^(2) phi` [sum of an infinite GP] `= 1/((1-sin^(2) phi))=1/(cos^(2) phi)`. `:. cos^(2) phi =1/y` ...(ii) `z= sum_(n=0)^(oo) cos^(2n) theta sin^(2n) phi =1+ cos^(2) theta sin^(2) phi +cos^(4) theta sin^(4) phi +...oo` `=a/((1-r))`, where `a=1` and `r=cos^(2) theta sin^(2) phi` `1/((1-cos^(2) theta sin^(2) phi))=1/(1-(1-sin^(2) theta)(1- cos^(2) phi))` `=1/(1-(1-1/x)(1-1/y))=1/(1-(1-1/x-1/y+1/(xy)))` [using (i) and (ii)] `=1/((1/x+1/y+1/(xy)))=(xy)/((x+y-1))` `:. z=(xy)/(x+y-1) implies xz+yz-z=xy`. Hence, `xz+yz-z=xy`.