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Question

If 0 < φ < (π/2), x= displaystyle∑n=0∞ cos 2 n φ, y= displaystyle∑n=0∞ sin 2 n φ and z= displaystyle∑n=0∞ cos 2 n φ sin 2 n φ, then (A) x y z=x z+y (B) x y z=x y+z (C) x y z=x+y+z (D) x y z=y z+x

Tardigrade Admin · Accepted Answer

Since, x= displaystyle∑n=0∞ cos 2 n φ =1+ cos 2 φ+ cos 4 φ+ ldots =(1/1- cos 2 φ)=(1/ sin 2 φ) (∵| cos x|<1) Similarly, y=(1/1- sin 2 φ)=(1/ cos 2 φ) and z=(1/1- sin 2 φ cos 2 φ)=(1/1-(1)x ⋅ (1)y=(x y/x y-1) ⇒ x y z=x y+z

Q. If $0 < ϕ < \frac{π}{2}, x = n = 0 \sum \infty cos^{2 n} ϕ, y = n = 0 \sum \infty sin^{2 n} ϕ$ and $z = n = 0 \sum \infty cos^{2 n} ϕ sin^{2 n} ϕ$ , then

$x yz = x z + y$

$x yz = x y + z$

$x yz = x + y + z$

$x yz = yz + x$

Q. If 0<ϕ<2π​,x=n=0∑∞​cos2nϕ,y=n=0∑∞​sin2nϕ and z=n=0∑∞​cos2nϕsin2nϕ, then

xyz=xz+y

xyz=xy+z

xyz=x+y+z

xyz=yz+x

Since, x=n=0∑∞​cos2nϕ =1+cos2ϕ+cos4ϕ+… =1−cos2ϕ1​=sin2ϕ1​(∵∣cosx∣<1) Similarly, y=1−sin2ϕ1​=cos2ϕ1​ and z=1−sin2ϕcos2ϕ1​=1−x1​⋅y1​1​=xy−1xy​ ⇒xyz=xy+z

Q. If $0 < ϕ < \frac{π}{2}, x = n = 0 \sum \infty cos^{2 n} ϕ, y = n = 0 \sum \infty sin^{2 n} ϕ$ and $z = n = 0 \sum \infty cos^{2 n} ϕ sin^{2 n} ϕ$ , then

$x yz = x z + y$

$x yz = x y + z$

$x yz = x + y + z$

$x yz = yz + x$