Trigonometric EquationVery Hard
Question
For 0 < φ < π/2, if x = 
cos2nφ, y =sin2nφ, z = cos2n φ sin2, then
cos2nφ, y =sin2nφ, z = cos2n φ sin2, thenOptions
A.xyz = xz + y
B.xyz = xy + z
C.xyz = x + y + z
D.xyz = yz + x
Solution
x =
cos2nφ = 1 + cos2 φ + cos4 φ + cos6 φ + .....
It is clearly a GP with common ration of cos2 φ which is ≤ 1
Hence, x =
Similarly, y =
and z =
Now, x + y =
Again,
= 1 - sin2 φ cos2 φ = 1 - 
⇒
⇒ xy = xyz - z
⇒ xy + z = xyz .....(i)
Therefore, (b) is the answer from Eq. (i) (putting the value of xy)
⇒ xyz = x + y + z
Therefore, (c) is also the answer.
cos2nφ = 1 + cos2 φ + cos4 φ + cos6 φ + .....It is clearly a GP with common ration of cos2 φ which is ≤ 1
Hence, x =
Similarly, y =
and z =
Now, x + y =
Again,
= 1 - sin2 φ cos2 φ = 1 - ⇒
⇒ xy = xyz - z
⇒ xy + z = xyz .....(i)
Therefore, (b) is the answer from Eq. (i) (putting the value of xy)
⇒ xyz = x + y + z
Therefore, (c) is also the answer.
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