Question

for 0<θ<π/2,if x=∑n=0 to ∞ sin^2nΦ,y=∑n=0 to ∞ cos^2nΦ and z=∑n=0 to ∞ cos^2nΦ sin^2nΦ ,then

Answer 1

x=1+sin^2x+sin^4x +sin^6x +,,,........
use now ,

x=1/(1-sin^2x)=sec^2x
in the same way
y=1/(1-cos^2x)=cosec^2x
x+y=sec^2x+cosec^2x
=sec^2x.cosec^2x=xy



z=1+sin^2x.cos^2x +sin^4x.cos^4x +....
z=1/(1-sin^2x.cos^2x)
=1/(1-1/xy)
=xy/(xy-1)

xyz-z=xy

also
x+y+z=xyz

Answer 2

It is not clear whether sin²ⁿ  Ф   or  sin²  nФ  ...

As  0 <Ф <π/2,   0 < sin² Ф < 1  ,   0 < cos² Ф < 1,  and also
                           0 <  sin² Ф * cos² Ф < 1

GP:  1 + r + r² + r³ + r⁴ + .... = 1/(1-r)    if  0 < r < 1

x = 1 + Sin² Ф + sin⁴ Ф + Sin⁶ Ф + .....∞
   = 1 /(1 - Sin² Ф) = Sec² Ф
=> 1/x = Cos² Ф

y = 1 + Cos² Ф + Cos⁴ Ф + Cos⁶ Ф + ... ∞
   = 1 /(1 - Cos² Ф) = Cosec² Ф
 => 1/y = Sin² Ф

z = 1 + Cos² Ф Sin² Ф  + Cos⁴ Ф Sin⁴ Ф + Cos⁶ Ф Sin⁶ Ф + ... ∞
   = 1 / (1 - Cos² Ф Sin² Ф)
   = 1 / (1 -  1/xy)

So  1/xy  + 1/z  = 1           or,   z =  xy / (xy - 1)        or,  xy = z /(z-1)