If `0 lt theta lt pi/2, x= underset(n=0)overset(oo)sum cos^(2n) theta, y= underset(n=0)overset(oo) sumsin^(2n) theta` and `z=underset(n=0)overset(oo)sum cos^(2n) theta* Sin^(2n) theta`, then show `xyz=xy+z`.

Options

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

Answer

B, C

Explanation

`:. 0ltphilt(pi)/(2)`
`:.0sinphi lt1` and `0ltcosphilt1`
`:.x=sum_(n=0)^(oo)cos^(2n) phi=1+cos^(2)phi+cos^(4)phi+"....."+oo`
`=(1)/(1-cos^(2)phi)=(1)/(sin^(2)phi)`
or `sin^(2)phi=(1)/(x)" " "…..(i)"`
and `y=sum_(n=0)^(oo)sin^(2n) phi=1+sin^(2)phi+sin^(4)phi+"....."+oo`
`=(1)/(1-sin^(2)phi)=(1)/(cos^(2)phi)`
or `cos^(2)phi=(1)/(y)" " ".....(ii)"`
From Eqs. (i) and (ii),
`sin^(2)phi+cos^(2)phi=(1)/(x)+(1)/(y)`
`1=(1)/(x)+(1)/(y)`
`:.xy=x+y" " "..........(iii)"`
and `z=sum_(n=0)^(oo)cos^(2n) phisin^(2)phi`
`=1+cos^(2)phisin^(2)phi+cos^(4)phisin^(4)phi+"......."`
`(1)/(1-sin^(2)phicos^(2)phi)=(1)/(1-(1)/(xy))[" from Eqs. (i)and (ii) "]`
`implies z=(xy)/(xy-1)`
`implies xyz=+xy`
and `xyz=z+x+y" " ["from Eq.(iii) "]`