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Question

If 0<θ, φ<(π/2), x = undersetn=0 overset∞ Sigma cos 2 n θ, y = undersetn=0 overset∞ Sigma sin 2 n φ and z = undersetn=0 overset∞ Sigma cos 2 n θ ⋅ sin 2 n φ then : (A) x y-z=(x+y) z (B) x y+y z+z x=z (C) x y z=4 (D) x y+z=(x+y) z

Tardigrade Admin · Accepted Answer

x=(1/1- cos 2 θ) ⇒ sin 2 θ=(1/x) Also, cos 2 θ=(1/y) 1- sin 2 θ cos 2 θ=(1/z) So, 1-(1/x) × (1/y)=(1/z) ⇒ z(x y-1)=x y ldots (1) Also, (1/x)+(1/y)=1 ⇒ x+y=x y dots (2) From (i) and (ii) xy + z = xyz =( x + y ) z

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

$x y - z = (x + y) z$

$x y + yz + z x = z$

$x yz = 4$

$x y + z = (x + y) z$

Q. If 0<θ,ϕ<2π​,x=n=0Σ∞​cos2nθ,y=n=0Σ∞​sin2nϕ and z=n=0Σ∞​cos2nθ⋅sin2nϕ then :

xy−z=(x+y)z

xy+yz+zx=z

xyz=4

xy+z=(x+y)z

x=1−cos2θ1​ ⇒sin2θ=x1​ Also, cos2θ=y1​ & 1−sin2θcos2θ=z1​ So, 1−x1​×y1​=z1​⇒z(xy−1)=xy…(1) Also, x1​+y1​=1 ⇒x+y=xy…(2) From (i) and (ii) xy+z=xyz=(x+y)z

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

$x y - z = (x + y) z$

$x y + yz + z x = z$

$x yz = 4$

$x y + z = (x + y) z$